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2 2.1
LIMITS AND DERIVATIVES The Tangent and Velocity Problems
1. (a) Using
(15 250), we construct the following table:
(b) Using the values of that correspond to the points closest
( = 10 and = 20), we have
slope = 5
(5 694)
694−250 5−15
= − 444 10 = −44 4
10
(10 444)
444−250 10−15
= − 194 5 = −38 8
20
(20 111)
111−250 20−15
= − 139 5 = −27 8
25
(25 28)
28−250 25−15
30
(30 0)
0−250 30−15
−38 8 + (−27 8) = −33 3 2
= − 222 10 = −22 2
= − 250 15 = −16 6
(c) From the graph, we can estimate the slope of the tangent line a
to be
2. (a) Slope =
2948 − 2530 42 − 36
(c) Slope =
2948 − 2806 42 − 40
−300 9
=
418 6
=
142 2
= −33 3.
≈ 69 67 = 71
(b) Slope =
2948 − 2661 42 − 38
(d) Slope =
3080 − 2948 44 − 42
=
287 4
= 71 75
=
132 2
= 66
From the data, we see that the patient’s heart rate is decreasing from 71 to 66 heartbeats minute after 42 minutes. After being stable for a while, the patient’s heart rate is dropping. 1 3. (a) = (b) The slope appears to be 1. , (2 −1) 1− (c) Using = 1, an equation of the tangent line to the ( 1 (1 − )) curve at(2 (i)
15
(1 5 −2)
2
(ii)
19
(1 9 −1 111 111)
1 111 111
1) is − =
(= 1( 1)
2), or − −
−
− 3.
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67
68
¤
4. (a)
CHAPTER 2
LIMITS AND DERIVATIVES
= cos
(0 5 0)
,
(b) The slope appears to be − . − 0 = − ( − 0 5) or
(c)
5. (a)
(i)
0
(0 1)
−2
(ii)
04
(0 4 0 309017)
−3 090170
(iii)
0 49
(0 49 0 031411)
−3 141076
(iv)
0 499
(0 499 0 003142)
−3 141587
(v)
1
(1 −1)
−2
(vi)
06
(0 6 −0 309017)
−3 090170
(vii)
0 51
(0 51 −0 031411)
−3 141076
(viii)
0 501
(0 501 −0 003142)
−3 141587
= ( ) = 40 − 16 2 . At = 2,
ave
=
(i) [2 2 5]:
= 0 5,
(iii) [2 2 05]:
ave
= 0 05,
= −32 ft s = −24 8 ft s
= ( ) = 10 − 1 86 2 . At = 1,
ave
=
(1 + ) − 1
(i) [1 2]:
= 1,
(iii) [1 1 1]: (v) [1 1 001]:
ave
= 0 1,
(iv) [2 2 01]:
ave
= 0 01,
6 28 − 1 86
(ii) [1 1 5]:
= 6 094 m s ave
(iv) [1 1 01]:
= 0 5,
2
(4) − (3) =
(iii) On the interval [4 5] ,
ave
=
(iv) On the interval [4 6] ,
ave
=
4− 3
79 2 − 46 5 =
(5) − (4) 5− 4
= 32 7 ft s.
124 8 − 79 2 =
(6) − (4) 6− 4
1
1
= 45 6 ft s.
176 7 − 79 2 =
2
ave
= 0 01,
(b) The instantaneous velocity when = 1 approaches 0) is 6 28 m s. (4) − (2) 79 2 − 20 6 = 7. (a) (i) On the interval [2 4] , ave = = 29 3 ft s. 4− 2 2
ave
= −24 16 ft s
= 6 28 − 1 86 , if
= 6 27814 m s
(ii) On the interval [3 4] ,
= 0.
= −25 6 ft s
ave
=
= 4 42 m s
= 0 001,
= 0 1,
if
= 10(1) − 1 86(1)2 = 8 14. The average velocity between times 1 and 1 +
=
ave
is
approaches 0) is −24 ft s.
10(1 + ) − 1 86(1 + )2 − 8 14
(1 + ) − (1)
.
2
= −24 − 16
(ii) [2 2 1]:
(b) The instantaneous velocity when = 2 6. (a)
−24 − 16 =
ave
1 2
= 40(2) − 16(2)2 = 16. The average velocity between times 2 and 2 +
=
(2 + ) − 2
+
(d)
40(2 + ) − 16(2 + )2 − 16
(2 + ) − (2)
=−
= 48 75 ft s.
= 5 35 m s
ave
= 6 2614 m s
= 0.
is
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SECTION 2.1
THE TANGENT AND VELOCITY PROBLEMS
¤
(b) Using the points (2 16) and (5 105) from the approximate tangent line, the instantaneous velocity at = 3 is about 105 − 16
=
5− 2
89 ≈
3
29 7 ft s.
(2) − (1) 8. (a) (i)
= ( ) = 2 sin
+ 3 cos
. On the interval [1 2], (1 1) − (1)
(ii) On the interval [1 1 1],
ave
=
ave
=
ave
=
=
= 6 cm s.
1
1 01 − 1
1 001 − 1
= −4 71 cm s.
−3 0613 − (−3) ≈
= −6 13 cm s.
0 01
(1 001) − (1) (iv) On the interval [1 1 001],
2− 1
01
(1 01) − (1) (iii) On the interval [1 1 01],
=
−3 471 − (−3) ≈
11− 1
ave
3 − (−3)
−3 00627 − (−3) ≈
= −6 27 cm s.
0 001
(b) The instantaneous velocity of the particle when = 1 appears to be about −6 3 cm s. 9. (a) For the curve
As
= sin(10
) and the point
(1 0):
2
(2 0)
0
05
(0 5 0)
0
15
(1 5 0 8660)
1 7321
06
(0 6 0 8660)
−2 1651
14
(1 4 −0 4339)
−1 0847
07
(0 7 0 7818)
−2 6061
13
(1 3 −0 8230)
−2 7433
08
(0 8 1)
−5
12
(1 2 0 8660)
4 3301
09
(0 9 −0 3420)
11
(1 1 −0 2817)
3 4202
−2 8173
approaches 1, the slopes do not appear to be approaching any particular value.
(b)
We see that problems with estimation are caused by the frequent oscillations of the graph. The tangent is so steep at
that we need to
take -values much closer to 1 in order to get accurate estimates of its slope.
(c) If we choose
= 1 001, then the poin
(0 999 0 0314) and tangent line a
is (1 001 −0 0314) and
≈ −31 3794. If
= 0 999, the
is
= −31 4422. The average of these slopes is −31 4108. So we estimate that the slope of the
is about −31 4.
69
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70
2.2
¤
CHAPTER 2
LIMITS AND DERIVATIVES
The Limit of a Function
1. As
approaches 2, ( ) approaches 5. [Or, the values of ( ) can be made as close to 5 as we like by taking
close to 2 (but 2. As
sufficiently
= 2).] Yes, the graph could have a hole at (2 5) and be defined such that (2) = 3.
approaches 1 from the left, ( ) approaches 3; and as
approaches 1 from the right, ( ) approaches 7. No, the limit
does not exist because the left- and right-hand limits are different. ( ) = ∞ means that the values of ( ) can be made arbitrarily large (as large as we please) by taking →−3
3. (a) lim
sufficiently close to −3 (but not equal to −3). (b) lim+ ( ) = −∞ means that the values of ( ) can be made arbitrarily large negative by taking →4
sufficiently close to 4
through values larger than 4. 4. (a) As
approaches 2 from the left, the values of ( ) approach 3, so lim
→2−
(b) As
( ) = 3.
approaches 2 from the right, the values of ( ) approach 1, so lim
( ) = 1.
→2+
(c) lim ( ) does not exist since the left-hand limit does not equal the right-hand limit. →2
(d) When (e) As
= 2,
= 3, so (2) = 3.
approaches 4, the values of ( ) approach 4, so lim ( ) = 4. →4
(f ) There is no value of ( ) when = 4, so (4) does not exist. 5. (a) As approaches 1, the values of ( ) approach 2, so lim ( ) = 2. →1
(b) As
approaches 3 from the left, the values of ( ) approach 1, so lim
(c) As
approaches 3 from the right, the values of ( ) approach 4, so lim
→3−
( ) = 1. ( ) = 4.
→3+
(d) lim ( ) does not exist since the left-hand limit does not equal the right-hand limit. →3
(e) When
= 3,
= 3, so (3) = 3. approaches −3 from the left, so
6. (a) ( ) approaches 4 as
(b) ( ) approaches 4 as
lim
→−3−
approaches −3 from the right, so
( ) = 4.
lim
( ) = 4.
→−3+
(c) lim
→−3
( ) = 4 because the limits in part (a) and part (b) are equal.
(d) (−3) is not defined, so it doesn’t exist. (e) ( ) approaches 1 as (f ) ( ) approaches −1 as
approaches 0 from the left, so lim
→0−
( ) = 1.
approaches 0 from the right, so lim
→0+
( ) = −1.
(g) lim ( ) does not exist because the limits in part (e) and part (f ) are not equal. →0
(h) (0) = 1 since the point (0 1) is on the graph of . (i) Since lim
→2−
( ) = 2 and lim+ ( ) = 2, we have lim ( ) = 2. →2
→2
(j) (2) is not defined, so it doesn’t exist. ° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.2
(k) ( ) approaches 3 as
approaches 5 from the left, so lim
→5−
( ) = −1
→0−
¤
→5
(l) ( ) does not approach any one number as
7. (a) lim
THE LIMIT OF A FUNCTION
approaches 5 from the right, so lim+ ( ) = 3.
(b) lim
→0+
( ) does not exist.
( ) = −2
(c) lim ( ) does not exist because the limits in part (a) and part (b) are not equal. →0
(d) lim− ( ) = 2
(e) lim+ ( ) = 0
→2
→2
(f ) lim ( ) does not exist because the limits in part (d) and part (e) are not equal. →2
(g) (2) = 1 8. (a) lim
→−3
(h) lim ( ) = 3 →4
( )=∞
(b) lim− ( ) = −∞ →2
(c) lim+ ( ) = ∞
( ) = −∞
(d) lim
→−1
→2
(e) The equations of the vertical asymptotes are = −3, = −1 and ( ) = −∞ (b) lim ( ) = ∞
9. (a) lim
→−7
(d) lim
→6−
10.
→−3
( ) = −∞
(e) lim
→6+
= 2. (c) lim ( ) = ∞ →0
( )= ∞
(f ) The equations of the vertical asymptotes are = −7, = −3, = 0, and = 6. lim ( ) = 150 mg and lim ( ) = 300 mg. These limits show that there is an abrupt change in the amount of drug in →12−
+
→12
the patient’s bloodstream at = 12 h. The left-hand limit represents the amount of the drug just before the fourth injection. The right-hand limit represents the amount of the drug just after the fourth injection. 11. From the graph of
1+ ( )=
2
2−
−1
if
if −1 ≤ ≥1
if
we see that lim ( ) exists for all
except
right and left limits are different at
= −1.
→
1,
= −1. Notice that the
12. From the graph of
1 + sin ( ) = cos sin
if if 0 ≤
≤
,
if
we see that lim ( ) exists for all
except
right and left limits are different at
= .
→
0
= . Notice that the
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71
72
¤
CHAPTER 2
13. (a) lim
→0−
LIMITS AND DERIVATIVES
( )=1
(b) lim+ ( ) = 0 →0
(c) lim ( ) does not exist because the limits →0
in part (a) and part (b) are not equal. ( ) = −1
14. (a) lim
→0−
(b) lim+ ( ) = 1 →0
(c) lim ( ) does not exist because the limits →0
in part (a) and part (b) are not equal. ( ) = −1, lim ( ) = 2, (0) = 1
15. lim
→0−
16. lim
( ) = 1, lim
→0
→0+
( ) = −2, lim
→3−
( ) = 2,
→3+
(0) = −1, (3) = 1
17. lim+ ( ) = 4, →3
(3) = 3,
( ) = 2, lim
→−2
( ) = 2,
2
2
−3
→0−
( ) = 2, lim+ ( ) = 0, lim →0
→4−
( ) = 3,
→4
:
−9 ( ) ( )
0 508 197 0 504 132
29
0 491 525
3 01
0 500 832
2 95
0 495 798
3 001
0 500 083
2 99
0 499 165
3 0001
0 500 008
2 999
0 499 917
2 9999
0 499 992
3 05
18. lim
lim+ ( ) = 0, (0) = 2, (4) = 1
(−2) = 1
19. For ( ) =
31
lim
→3−
1 −3 = . 2 −9
2
It appears that lim
→3
2
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.2
−3
2
20. For ( ) =
¤
:
−9
2
( )
( )
−2 5
−5
−3 5
7
−2 9
−29
−3 1
31
−2 95
−59
−3 05
61
−2 99
−299
−3 01
301
−2 999
−2999
−3 001
3001
−2 9999
THE LIMIT OF A FUNCTION
−29,999
−3 0001
It appears that
( ) = −∞ and that
lim
→−3+
( ) = ∞, so lim
lim
2
−3
30,001 →−3
→−3−
2
does not exist.
−9
5 5
21. For ( ) =
−1
22. For ( ) =
:
(2 + ) − 32
:
( )
( )
05
131 312 500
−0 5
48 812 500
05
22 364 988
−0 5
1 835 830
01
88 410 100
−0 1
72 390 100
01
6 487 213
−0 1
3 934 693
0 01
80 804 010
−0 01
79 203 990
0 01
5 127 110
−0 01
4 877 058
0 001
80 080 040
−0 001
79 920 040
0 001
5 012 521
−0 001
4 987 521
0 0001
80 008 000
−0 0001
79 992 000
0 0001
5 001 250
−0 0001
4 998 750
()
()
5
It appears that lim
−1
It appears that lim
= 5.
→0
23. For ( ) =
→0
− ln 4 : −4
ln
( )
( )
39
0 253 178
41
0 246 926
3 99
0 250 313
4 01
0 249 688
3 999
0 250 031
4 001
0 249 969
3 9999
0 250 003
4 0001
0 249 997
It appears that lim ( ) = 0 25. The graph confirms that result. →4
24. For ( ) =
1+ 1+
9 15 :
( )
( )
−1 1
0 427 397
−0 9
0 771 405
−1 01
0 582 008
−0 99
0 617 992
−1 001
0 598 200
−0 999
0 601 800
−1 0001
0 599 820
−0 9999
0 600 180
It appears that lim
→−1
( ) = 0 6. The graph confirms that result.
(2 + )5 − 32
= 80.
73
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74
¤
CHAPTER 2
25. For ( ) =
LIMITS AND DERIVATIVES
sin 3 : tan 2 ( )
±0 1
It appears that lim
1 457 847
±0 01
1 499 575
±0 001
1 499 996
±0 0001
1 500 000
26. For ( ) =
5 −1
→0
sin 3
= 1 5.
tan 2
The graph confirms that result.
: ()
()
−0 1
1 486 601
01
1 746 189
−0 01
1 596 556
0 01
1 622 459
−0 001
1 608 143
0 001
1 610 734
−0 0001
1 609 308
0 0001
1 609 567
It appears that lim ( ) ≈ 1 6094. The graph confirms that result. →0 27. For ( ) =
: It appears that lim+ ( ) = 1.
( ) 01
0 794 328
0 01
0 954 993
0 001
0 993 116
0 0001
0 999 079
28. For ( ) =
2
→0
The graph confirms that result.
ln : It appears that lim+ ( ) = 0.
( ) 01
−0 023 026
0 01
−0 000 461
0 001
−0 000 007
0 0001
−0 000 000
→0
The graph confirms that result.
29. (a) From the graphs, it seems that lim
→0
cos 2 − cos 2
= −1 5.
(b) ( ) ±0 1
−1 493 759
±0 01
−1 499 938
±0 001
−1 499 999
±0 0001
−1 500 000
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SECTION 2.2
30. (a) From the graphs, it seems that lim →0
sin
= 0 32.
¤
THE LIMIT OF A FUNCTION
75
(b) ( )
sin ±0 1
0 323 068
±0 01
0 318 357
±0 001 ±0 0001
0 318 310 0 318 310
Later we will be able to show that 1 the exact value is . +1
31. lim
−5 +1
→5+
32. lim →5−
33. lim →1
= ∞ since the numerator is positive and the denominator approaches 0 from the positive side as
→ 5+ .
= −∞ since the numerator is positive and the denominator approaches 0 from the negative side as
−5 2−
→ 5− .
= ∞ since the numerator is positive and the denominator approaches 0 through positive values as
→ 1.
( − 1)2 √
34. lim− →3
( − 3)5 = −∞ since the numerator is positive and the denominator approaches 0 from the negative side as 2
35. Let =
− 9. Then as
→0
→(
38.
1
lim
→ − −
→
lim
→0+
→ 0+ .
is positive and sec
→ −∞ as
→ (
2)+ .
csc
sin
= −∞ since the numerator is positive and the denominator approaches 0 through negative
= lim →2 −
→ 2
values as 2
40. lim
−
−2
−4 +4
2
= −∞ since the numerator is negative and the denominator approaches 0 through positive values
.
→2 −
→2−
cos
= lim
→ −
39.
1
→ 0+ as
− 9) = lim ln = −∞ by (5).
2)+
lim cot
as
= −∞ since
sec
2
→3+
) = −∞ since sin
36. lim+ ln(sin
37.
→ 3+ , → 0+ , and lim ln(
→ 3− .
sin
. ( − 2)
= lim →2−
( − 2)2
= −∞ since the numerator is positive and the denominator
= lim →2−
−2
approaches 0 through negative values as → 2− . 2 ( − 4)( + 2) −2 −8 41. lim = ∞ since the numerator is negative and the denominator approaches 0 through = lim →2+
2
−5 +6
negative values as 42. lim
→0+
1
− ln
→2+
( − 3)( − 2)
→ 2+ . = ∞ since
1
→ ∞ and ln
→ −∞ as
→ 0+ .
43. lim (ln →0
2
−
−2
) = −∞ since ln
2
→ −∞ and
−2
→ ∞ as
→ 0.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
76
¤
CHAPTER 2
LIMITS AND DERIVATIVES 2
44. (a) The denominator of
=
3 −2 = 0 and
=
3 2
2
+1
=
+1
is equal to zero when
(b)
(3 − 2 )
2
(and the numerator is not), so
= 0 and
= 1 5 are
vertical asymptotes of the function.
45. (a) ( ) =
3
1 . −1
From these calculations, it seems that lim
→1−
(b) If
( ) = −∞ and lim+ ( ) = ∞. →1
3
is slightly smaller than 1, then
→1−
is slightly larger than 1, then
3
−1 14 −3 69 −33 7 −333 7 −3333 7 −33,333 7
− 1 will be a negative number close to 0, and the reciprocal of
will be a negative number with large absolute value. So lim
If
15 11 1 01 1 001 1 0001 1 00001
( ) 05 09 0 99 0 999 0 9999 0 99999
( ) 0 42 3 02 33 0 333 0 3333 0 33,333 3
3
− 1, that is, ( ),
( ) = −∞.
− 1 will be a small positive number, and its reciprocal, ( ), will be a large positive
number. So lim+ ( ) = ∞. →1
(c) It appears from the graph of lim
→1−
that
( ) = −∞ and lim
→1+
( ) = ∞.
46. (a) From the graphs, it seems that lim
→0
tan 4
= 4.
(b) ±0 1 ±0 01 ±0 001 ±0 0001
47. (a) Let ( ) = (1 + )
−0 001 −0 0001 −0 00001 −0 000001 0 000001 0 00001 0 0001 0 001
1
( ) 4 227 932 4 002 135 4 000 021 4 000 000
. ( ) 2 71964 2 71842 2 71830 2 71828 2 71828 2 71827 2 71815 2 71692
(b)
It appears that lim (1 + )1 →0
≈ 2 71828, which is approximately .
In Section 3.6 we will see that the value of the limit is exactly .
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.2
THE LIMIT OF A FUNCTION
¤
77
48. (a)
No, because the calculator-produced graph of ( ) =
= 4. A second graph, obtained by increasing the numpoints option in Maple, begins to
has an infinite discontinuity at reveal the discontinuity at
+ ln − 4 looks like an exponential function, but the graph of
= 4.
(b) There isn’t a single graph that shows all the features of . Several graphs are needed since negative values of
and like
for
5, but yet has the infinite discontiuity at
looks like ln − 4 for large
= 4.
A hand-drawn graph, though distorted, might be better at revealing the main features of this function.
2
49. For ( ) =
− (2 1000): (b)
(a) 1 08
( ) 0 998 000 0 638 259
06 04 02 01 0 05
0 358 484 0 158 680 0 038 851 0 008 928 0 001 465
0 04 0 02 0 01 0 005 0 003 0 001
( ) 0 000 572 −0 000 614 −0 000 907 −0 000 978 −0 000 993 −0 001 000
It appears that lim ( ) = −0 001. →0
It appears that lim ( ) = 0. →0
50. For ( ) =
tan
− 3
:
(a) 10 05 01 0 05 0 01 0 005
( ) 0 557 407 73 0 370 419 92 0 334 672 09 0 333 667 00 0 333 346 67 0 333 336 67
(b) It seems that lim ( ) = 1 . →0
3
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
78
¤
CHAPTER 2
LIMITS AND DERIVATIVES
(c)
Here the values will vary from one ( ) 0 001 0 0005 0 0001 0 00005 0 00001 0 000001
calculator to another. Every calculator will
0 333 333 50 0 333 333 44 0 333 330 00 0 333 336 00 0 333 000 00 0 000 000 00
eventually give false values.
(d) As in part (c), when we take a small enough viewing rectangle we get incorrect output.
51. No matter how many times we zoom in toward the origin, the graphs of ( ) = sin(
→ 0.
lines. This indicates more and more frequent oscillations as
52. (a) For any positive integer , if
=
1
, then ( ) = tan
) appear to consist of almost-vertical
1
= tan(
) = 0. (Remember that the tangent function has
period .)
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.3
(b) For any nonnegative number , if
( ) = tan
1
CALCULATING LIMITS USING THE LIMIT LAWS
= tan
(4 + 1)
= tan
4
+
+
= tan
4
= tan
4
does not exist since ( ) does not get close to a fixed number as
4
must have 2 sin −1 ≤ sin
=
2
+
, or equivalently, sin
≤ 1, we must have sin
−1
. So
−1
− sin
=±
→ −
3
55. (a) Let
=
√
1−
−1
2
−
,
1−
as
approaches 1 is 6.
3
(b) We need to have 5 5
4
(corresponding
− sin−1
4
is the
are also equations of
≈ ±2 24).
→ 0+ , and
→ ∞.
−1
√
0 99 0 999 0 9999 1 01 1 001 1 0001
5 925 31 5 992 50 5 999 25 6 075 31 6 007 50 6 000 75
6 5. From the graph we obtain the approximate points of intersection
(0 9314 5 5)
−1
(1 0649 6 5). Now 1 − 0 9314 = 0 0686 and 1 0649 − 1 = 0 0649, so by requiring that
we ensure that
2.3
2
. Since
2
.
From the table and the graph, we guess
and
2
. Thus, we
2
−1
that the limit of
+
+
4
vertical asymptotes (corresponding to
→ −
4
2
= ± sin−1
= ± 4 and so
4
→
=
=
≈ ±0 90). Just as 150◦ is the reference angle for 30◦ ,
reference angle for sin
. As
≈ ±0 90
= tan(2 sin ) at
graph of the tangent function has vertical asymptotes at
0
1
≈ ±2 24. To find the exact equations of these asymptotes, we note that the
and
= lim
tan → 0. Thus, lim →0
→ 0.
There appear to be vertical asymptotes of the curve
54. lim
=1
4
(c) From part (a), ( ) = 0 infinitely often as → 0. From part (b), ( ) = 1 infinitely often as
to
79
4 , then (4 + 1)
=
4
53.
¤
be within 0 0649 of 1,
is within 0 5 of 6.
Calculating Limits Using the Limit Laws
1. (a) lim [ ( ) + 5 ( )] = lim →2
→2
( ) + lim [5 ( )] →2
= lim ( ) + 5 lim ( ) →2
[Limit Law 1]
3
→2
[Limit Law 3]
3
(b) lim [ ( )] =
lim ( )
[Limit Law 6]
→2
= ( −2)3 = −8
→2
= 4 + 5(−2) = −6
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
80
¤
CHAPTER 2
(c) lim
LIMITS AND DERIVATIVES
( )=
→2
lim ( )
3 ( )
(d) lim
[Limit Law 11]
→2
→2
√ = 4=2
lim [3 ( )] →2
=
( )
[Limit Law 5]
lim ( ) →2
3 lim ( ) →2
=
lim ( ) →2
=
[Limit Law 3]
3(4) = −6 −2
(e) Because the limit of the denominator is 0, we can’t use Limit Law 5. The given limit, lim
→2
( ) , does not exist because the ( )
denominator approaches 0 while the numerator approaches a nonzero number. lim [ ( ) ( )] ( ) ( ) →2 (f ) lim = [Limit Law 5] →2
( )
lim ( ) →2
lim ( ) · lim ( ) →2
→2
=
[Limit Law 4]
lim ( ) →2
=
−2 · 0 =0 4
2. (a) lim [ ( ) + ( )] = lim →2
( ) + lim ( )
→2
[Limit Law 1]
→2
= −1 + 2 =1 (b) lim ( ) exists, but lim ( ) does not exist, so we cannot apply Limit Law 2 to lim [ ( ) − ( )]. →0
→0
→0
The limit does not exist. (c) lim [ ( ) ( )] = lim →−1
( ) · lim
→−1
( )
[Limit Law 4]
→−1
= 1 ·2 =2
(d) lim ( ) = 1, but lim ( ) = 0, so we cannot apply Limit Law 5 to lim →3 →3
→3
( )
Note: lim →3−
= ∞ since ( ) → 0+ as
→ 3− and lim
( )
→3+
( )
( )
( ) . The limit does not exist. ( )
= −∞ since ( ) → 0− as
→ 3+ .
Therefore, the limit does not exist, even as an infinite limit. 2
(e) lim
( ) = lim
→2
2
· lim ( )
→2
(f ) (−1) + lim
[Limit Law 4]
→2
( ) is undefined since (−1) is
→−1
2
not defined.
= 2 · (−1) = −4 3. lim (5 →3
3
−3
2
+
− 6) = lim (5
3
→3
= 5 lim
→3
) − lim (3
2
→3
3
− 3 lim
→3
) + lim
→3
2
+ lim
→3
− lim 6 →3
− lim 6 →3
[Limit Laws 2 and 1] [3]
= 5(33 ) − 3(32 ) + 3 − 6
[9, 8, and 7]
= 105
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.3
4.
4
lim ( →−1
− 3 )(
2
− 3 ) lim (
4
+ 5 + 3) = lim ( →−1
2
→−1
=
lim
4
=
lim
4
→−1
→−1
+ 5 + 3)
− lim 3
lim
2
− 3 lim
lim
2
→−1
→−1
[Limit Law 4]
3 + lim 5 + lim →−1 →−1
→−1
→−1
CALCULATING LIMITS USING THE LIMIT LAWS
+ 5 lim
→−1
+ lim 3 →−1
= (1 + 3)(1 − 5 + 3)
[2, 1]
[3]
[9, 8, and 7]
= 4(−1) = −4 4
5. lim →−2
2
−2
−3 +2
2
− 2)
4
lim (
→−2
=
[Limit Law 5] − 3 + 2)
2
lim (2
→−2
− lim 2
4
lim
→−2
=
− 3 lim
2
2 lim
→−2
→−2
[1, 2, and 3]
+ lim 2
→−2
→−2
16 − 2
=
[9, 7, and 8]
2(4) − 3(−2) + 2 = 6. lim
√
→−2
4
14 7 = 16 8
+3 +6 =
4
lim ( →−2
=
4
lim
+ 3 lim
[11]
+ lim 6 →−2
→−2
→−2
=
+ 3 + 6)
[1, 2, and 3]
4
(−2) + 3 (−2) + 6 √ √ = 16 − 6 + 6 = 16 = 4 7. lim (1 +
√ 3
) (2 − 6
2
3
+
) = lim (1 +
→8
√ 3
[9, 8, and 7]
) · lim (2 − 6
3
)
[Limit Law
→8
lim 1 + lim →8
= 1+
+
4]
→8
=
2
√ 3
· lim 2 − 6 lim
→8
→8
2
→8
√ 3 8 · 2 − 6 · 82 + 83
= (3)(130) = 390 2
8. lim →2
3
−2
2
−2
2
=
−3 +5
lim →2
3
→2 lim( 3 →2
2
lim
2
− 2)
[5]
− 3 + 5)
lim
=
[Limit Law 6]
−3 +5
lim(
=
2
→2 3
2
− lim 2 →2
3 lim + lim 5
2
[1, 2, and 3]
+ lim
3
[1, 2, and 3]
→8
[7, 10, 9]
¤
81
−
→2
= =
→2
4− 2 8 − 3(2) + 5 2 7
2
=
→2
2
[9, 7, and 8]
4 49
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
82
¤
CHAPTER 2 2
2
9. lim
+1
LIMITS AND DERIVATIVES
=
3 −2
→2
2
2
lim
+1
3 −2
→2
2
lim (2 →2
=
[Limit Law 11]
+ 1) [5]
lim (3 − 2) →2 2
2 lim
→2
=
+ lim 1 →2
→2
→2
2(2)2 + 1 = 3(2) − 2
=
[1, 2, and 3]
− lim 2
3 lim
9 3 = 4 2
[9, 8, and 7]
10. (a) The left-hand side of the equation is not defined for
(b) Since the equation holds for all
= 2, but the right-hand side is.
= 2, it follows that both sides of the equation approach the same limit as
in Example 3. Remember that in finding lim ( ), we never consider
= .
→
2
11. lim
−6 +5 −5
→5
2
12. lim →−3
2
13. lim
→5
2
+3
→4
2
lim
−
−4
→4+
2
15. lim →−3
2
16. lim →−1
17. lim
2
2
−4
( − 4)( + 3)
→4
=
−3 − 4
3 7
− 5 + 6 → 6 as
2
= lim
−3
→ 5.
= −∞ and
. The last limit does not exist since lim −4
→4−
−4
= ∞.
−9
+3 +1 −2 −3
(−5 + )2 − 25
→0
( + 3)( − 3)
= lim
+7 +3
2
2
→4
=
→−3
− 5 → 0, but
( + 3)
= lim
− 12
= lim
( − 4)( + 3)
→−3
− 5 +6 does not exist since −5
14. lim
→5
( + 3)
= lim
− 12
= lim ( − 1) = 5 − 1 = 4
−5
→5
+3
−
2
( − 5)( − 1)
= lim
→ 2, just as
→−3
(2 + 1)( + 1) ( − 3)( + 1)
→−1
(25 − 10 +
= lim →0
(2 + )3
→−3
(2 + 1)( + 3)
= lim
18 .l
−3
= lim
2
2 +1
=
= lim
=
−10 +
−6
=
−5
2(−1) + 1
−3
→−1
→0
im
−3 − 3 2(−3) + 1
2 +1
= lim
) − 25
=
=
−1 − 3
5 −1 −4
2
= lim →0
6
=
1 4
(−10 + )
= lim (−10 + ) = −10 →0
−8
8 + 12 +
= lim
6
2
+
3
12 + 6 2+
= lim
3
−8 →0
→0
→0
= lim 12 + 6 + →0
2
= 12 + 0 + 0 = 12
19. By the formula for the sum of cubes, we have
+2
lim →−2
3
+8
+2
= lim →−2
( + 2)(
2
− 2 + 4)
1
= lim →−2
2
−2 +4
=
1 4+ 4+4
=
1
.
12
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.3
CALCULATING LIMITS USING THE LIMIT LAWS
20. We use the difference of squares in the numerator and the difference of cubes in the denominator. 4
lim →1
21. lim
3
−1
(
= lim
−1
→1
√ 9+
−3
2
− 1)(
( − 1)(
+ 1)
= lim
−3
+3
9+
+3
→0
+3
2
2
+ 1)
→1
+ + 1)
→0
2
9+
1
9+
− 32
+ 1)
=
2(2)
+ +1
=
3
4 3
(9 + ) − 9 √
= lim →0
9+
+3
−
√ 1− √
2
+
1−
1
=
3+ 3
+3
2
+3
1
=
2
( + 1)(
= lim
√ 9+ √ = lim
= lim √
√ 9+
→0
( − 1)(
√ 9+ ·√
→0
= lim
( − 1)( + 1)(
→1
+ + 1)
√ 9+
= lim
→0
2
2
6 2
22. lim
√ 4 +1 − 3
= lim
−2
→2
−2
→2
( − 2)
− 32
4 +1+3
− 2)
4
= lim
√ ( − 2) 4 + 1 + 3
→2
→2
4 +1 + 3
4 +1 − 9
= lim
√ 4 +1 √
√ 4 +1 + 3 ·√ = lim
√ 4 +1 − 3
√ ( − 2) 4 + 1 + 3
→2
4 4 2 = lim √ = √ = →2 3 4 +1 + 3 9+3 1
−
23. lim →3
24. lim
1 1 1 − 3 = lim 3
−3
−3
→3
(3 + )−1 − 3−1
3 ·
→3
3
1 3 + = lim
→0
3−
= lim
−
= lim
3 ( − 3)
→0
1 →0
√ 1+
3(3 + )
√ 1+
=
→0
1+ (1 + ) − (1 − ) √ √ 1+ + 1− 2
= √
√ =
→0
1
1
− 2
+
= lim →0
1
−
2
(3 + )3
+
√ 1− √
+
1−
= lim →0
1
3(3 + 0)
=−
√ 1+ √ = lim →0
1+
2
9
2
= lim
√ √ 1+ + 1−
→0
=1
2
1+ 1 26. lim
√ 1−
−
1 =−
→0
→0
→0
9
→0
lim [3(3 + )]
= lim
= lim
3
1
−
25. lim
−
−
(3 + )3
√ 1+ ·√
−
√ 1−
→3
1
=
1 3 = lim 3 − (3 + ) = lim
→0
= lim −
−1
1 ( + 1)
= lim + 1 − 1 = lim →0
( + 1)
→0
1 +1
=
1 0+ 1
=1
2 √ 1+
+
√ 1−
¤
83
27. lim →16
4−
√
(4 − √ )(4 + √ )
= lim
16 −
→16
2
(16 − 1
= lim →16
2
28. lim →2
4
− 4 +4 −3
2
−4
4+
= lim →2
√
+
√
1
)
(
2
2
2
16
→2
= + 1)
=
0
1 16(8)
√
)
1
=
128
( − 2)2
= lim
+ 1)
−2 ( + 2)(
√
16 4 +
− 4)(
(16 − )(4 +
→16
)
=
( − 2)2
= lim →2
2 )(4
16 − = lim
( + 2)( − 2)(
2
+ 1)
=0
4 ·5
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
84
¤
CHAPTER 2
LIMITS AND DERIVATIVES
1 √ 1+
29. lim →0
−
√ 1+
1−
1
√
= lim →0
→0
30.
2
lim
√
+9 − 5
2
2
→−4
( + 4)
( + ) − 3
3
= lim
→0
2
3
(
2
2
3
+
(3
( + )2
2
2
2
+3
+
2
)
√
2
= lim
+9 +5
4
=−
5+ 5
3
2
= lim (3
− ( + )2
( + )2
= lim
→0
+9+ 5
5 2
3
2
+3
3
+
+3
+
2
)=3
2
→0
1 −
2
→0
→0
1
1
√
−8
=
→0
= lim
32. lim
)−
1+
+ 9) − 25
2
( + 4)
( + 4)
16 + 9 + 5
+3
1+
( + 4)( − 4)
→−4
−4 − 4
= √
√ 1+
2
1+ 0
→−4
= lim
+9 + 5
+3
=−
(
+9+5
+9+5
−4
= lim √ →−4
2
√
+9 +5
− 16
√
→0
= lim 2
( + 4)
= lim
31. lim
2
√
→−4
+4
= lim √
1+0 1+
√
−
1+
−1
= √
√ 1+ 1+
= lim
→−4
+1 1+
−1
√ 1+
1+ √
→0
1+
+9 − 5
√ 1+
√
= lim 1+
= lim √
√
1−
2
−(
2
= lim
→0
→0
2
+2
2(
+
2
− (2 + )
) = lim
+ )2
2(
+ )2
→0
= lim
→0
−(2 + ) −2 = 2 2 ( + )2 ·
33. (a)
2
=−
2 3
(b) ( )
lim √ →0
(c) lim
→0
1+ 3 − 1
≈
−0 001 −0 000 1 −0 000 01 −0 000 001 0 000 001 0 000 01 0 000 1 0 001
2 3
√ 1+ 3 +1 √ ·√ 1+ 3 − 1 1+3 +1
= lim
→0
0 666 166 3 0 666 616 7 0 666 661 7 0 666 666 2 0 666 667 2 0 666 671 7 0 666 716 7 0 667 166 3
√ 1+3 +1 = lim →0 (1 + 3 ) − 1
√ 1+3 +1 3
The limit appears to be
2 . 3
= =
√ 1 lim 1+3 +1 3 →0 1 3 1
=
3
[Limit Law 3]
lim (1 + 3 ) + lim 1 →0
→0
lim 1 + 3 lim →0
[1 and 11]
+1
[1, 3, and 7]
→0
=
1 √ 1 + 3· 0 + 1 3
=
1 2 (1 + 1) = 3 3
[7 and 8]
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.3
34. (a)
CALCULATING LIMITS USING THE LIMIT LAWS
¤
85
(b) ( )
lim
√ 3+
−
√ 3
→0
√ − 3
√ 3+
−0 001 −0 000 1 −0 000 01 −0 000 001 0 000 001 0 000 01 0 000 1 0 001
≈ 0 29
·√ 3+
(c) lim →0
The limit appears to be approximately 0 2887.
√ + 3
√ 3+
(3 + ) − 3
√ 3
+
0 288 699 2 0 288 677 5 0 288 675 4 0 288 675 2 0 288 675 1 0 288 674 9 0 288 672 7 0 288 651 1
√ = lim √ →0 3 3+
√
= lim →0
3+
1
+
+
√ 3
lim 1 √
= lim
→0
3+
→0
√
+ lim
→0
1
=
lim (3 + ) + →0
[Limit Laws 5 and 1]
3
√ 3
[7 and 11]
1 √ = √ 3+ 0+ 3 1 = √ 2 3 35. Let ( ) = −
2
−1 ≤ cos 20
, ( )=
2
≤1
−
⇒
cos 20 2
≤
2
and ( ) =
2
≤
2
cos 20
[1, 7, and 8]
. Then ⇒
( ) ≤ ( ) ≤ ( ).
So since lim ( ) = lim ( ) = 0, by the Squeeze Theorem we have →0
→0
lim ( ) = 0. →0
√
36. Let ( ) = −
3
−1 ≤ sin(
)≤ 1
2,
( )=
⇒
√ −
+
√
3
3
2
+
2
+
≤
sin(
√
3
), and ( ) = 2
+
sin(
)≤
√ √
3
3
+
+
2.
2
Then ⇒
( ) ≤ ( ) ≤ ( ). So since lim ( ) = lim ( ) = 0, by the Squeeze Theorem →0
→0
we have lim ( ) = 0. →0
37. We have lim (4 − 9) = 4(4) − 9 = 7 and lim →4
for
2
→4
− 4 + 7 = 42 − 4(4) + 7 = 7. Since 4 − 9 ≤ ( ) ≤
2
−4 +7
( ) = 7 by the Squeeze Theorem. ≥ 0, lim →4
38. We have lim (2 ) = 2(1) = 2 and lim ( →1
→1
4
−
2
+ 2) = 14 − 12 + 2 = 2. Since 2 ≤ ( ) ≤
4
−
2
lim ( ) = 2 by the Squeeze Theorem. →1
39. −1 ≤ cos(2
)≤ 1
⇒
−
4
≤
4
cos(2
)≤
4
. Since lim
4
= 0 and lim
4
= 0, we have
+ 2 for all ,
lim
→0
4
cos(2
) = 0 by the Squeeze Theorem.
→0
−
→0
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
86
¤
CHAPTER 2
40. −1 ≤ sin(
lim+ (
√
→0
LIMITS AND DERIVATIVES
)≤ 1
−1
⇒
≤
) = 0, we have lim+
sin(
√
≤
)
sin(
)
−3 −( − 3)
if
−3≥0
if
−3
0
−3
=
3−
Thus, lim (2 + − 3 ) = lim (2 + →3+
/ ≤
√
sin(
)
√
≤
. Since lim+ ( →0
√
/ ) = 0 and
= 0 by the Squeeze Theorem.
→0
41. − 3 =
√
⇒
1
if
≥3
if
3
− 3) = lim (3 − 3) = 3(3) − 3 = 6 and
→3+
→3+
lim (2 + − 3 ) = lim (2 + 3 − ) = lim ( + 3) = 3 + 3 = 6. Since the left and right limits are equal, →3−
→3−
→3−
lim (2 + − 3 ) = 6. →3
+6 42. + 6 =
+6 ≥0
if
−( + 6)
if
+6
0
+6
=
−( + 6)
if
≥ −6
if
−6
We’ll look at the one-sided limits. 2 + 12
lim
=
+ 6
→−6+
2
lim
+ 6)
= 2
and
+6
→−6+
→−6−
The left and right limits are different, so lim →−6
43. 2
3
−
2
Thus,
3
−
(2 − 1) =
2 −1
2 − 1 =
So 2
2
=
2
3
if 2 − 1
0
[−(2 − 1)] for
2 −1
lim →0 5−
2
=
−
= 2
2 + 12
2
lim
= −2
−( + 6)
→−6−
+ 6
+ 6)
does not exist.
=
2
2 − 1
2 −1
if
≥05
−(2 − 1)
if
05
0 5. 2 −1
lim →0 5−
=
+ 6
· 2 − 1 =
if 2 − 1 ≥ 0
−(2 − 1)
2
2
2 + 12
lim
2 [−(2
=
− 1)]
lim →0 5−
2− 44. Since = − for
0, we have lim
45. Since = − for
0, we have lim
→−2
→0−
1
→−2
−
−1
=
2
=
5)2
(0
1
→−2
1 − 1
= lim
2+
= lim
−
→0−
denominator approaches 0 and the numerator does not. 1 1 − = lim 46. Since = for 0, we have lim →0+ →0+
1
−
= −4.
2+ = lim
2+
−1 0 25
2 − (− ) = lim
2+
−1
1
= lim 1 = 1. →−2
2
, which does not exist since the
→0−
= lim 0 = 0. + →0
47. (a)
(b) (i) Since sgn (ii) Since sgn
= 1 for
→0
= −1 for
(iii) Since lim sgn →0−
0, lim+ sgn
= lim+ 1 = 1.
0, lim− sgn →0
= lim+ sgn , lim sgn
(iv) Since sgn = 1 for
→0
→0
→0
= lim −1 = −1. →0−
does not exist.
= 0, lim sgn = lim 1 = 1. →0
→0
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.3
−1 if sin 48. (a) ( ) = sgn(sin ) =
0
0
if sin
=0
1
if sin
0
(i) lim+ ( ) = lim+ sgn(sin
) = 1 since sin
(ii) lim
) = −1 since sin
→0
→0
( ) = lim sgn(sin −
→0−
→0
(iii) lim ( ) does not exist since lim →0
is positive for small positive values of .
) = −1 since sin
(v) lim
) = 1 since sin
→ −
( ) = lim sgn(sin → −
(vi) lim ( ) does not exist since lim →
( ).
→0−
(iv) lim+ ( ) = lim+ sgn(sin →
is negative for small negative values of .
( ) = lim
→0+
→
CALCULATING LIMITS USING THE LIMIT LAWS
is negative for values of is positive for values of
( ) = lim
( ).
(b) The sine function changes sign at every integer multiple of , so the signum function equals 1 on one side and −1 on the other side of an integer. Thus, lim ( ) does not exist for
=
→
2
( ) = lim
→2+
→2+
+
−6
,
an integer.
− 2
→2+
= lim+ →2
,
(c)
( + 3)( − 2)
= lim
− 2
slightly less than .
→ −
→ +
49. (a) (i) lim
slightly greater than .
( + 3)( − 2) −2
[since
−2
0 if
→ 2+ ]
= lim+ ( + 3) = 5 →2
(ii) The solution is similar to the solution in part (i), but now − 2 = 2 − Thus, lim
→2−
since
−2
( ) = lim −( + 3) = −5. →2−
(b) Since the right-hand and left-hand limits of at
=2
(c)
are not equal, lim ( ) does not exist. →2
2
50. (a) ( ) =
lim →1−
+1
if
1
( − 2)2
if
≥1
( ) = lim (
2
+ 1) = 12 + 1 = 2,
lim
→1−
(b) Since the right-hand and left-hand limits of are not equal, lim ( ) does not exist. →1
→1+
at
=1
( ) = lim ( − 2)2 = (−1)2 = 1 →1+
(c)
0 if
→ 2− .
¤
87
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88
¤
CHAPTER 2
51. For the lim
( ) to exist, the one-sided limits at = 2 must be equal. lim
→2
lim
→2+
LIMITS AND DERIVATIVES
→2−
√
( ) = lim
+ =
→2+
52. (a) (i) lim
→1−
(ii) lim
( ) = lim
→1−
√ 2+ .
√ 2+
⇒
9=2+
→2
⇔
) = 2 − 12 = 1. Since lim
2
( ) = 1 and lim
→1−
→1+
→1+
Note that the fact (1) = 3 does not affect the value of the limit. (iii) When (iv) lim
→2−
= 1, ( ) = 3, so (1) = 3. ( ) = lim− (2 −
2
→2
) = 2 − 22 = 2 − 4 = −2
(v) lim+ ( ) = lim+ ( − 3) = 2 − 3 = −1 →2
→2
(vi) lim ( ) does not exist since lim →2
( ) = lim
→2−
(b) 3 ( )=
2−
2
−3
if
1
if
=1
( ).
→2+
≤2
if 1 if
2
53. (a) (i) [[ ]] = −2 for −2 ≤
−1, so
(ii) [[ ]] = −3 for −3 ≤
−2, so
lim [ ]] =
→−2+
lim [[ ]] =
→−2−
lim (−2) = −2
→−2+
lim (−3) = −3.
→−2−
The right and left limits are different, so lim [[ ]] does not exist. →−2
(iii) [[ ]] = −3 for −3 ≤
−2, so
lim [[ ]] = 4
→−2
(b) (i) [[ ]] =
− 1 for
−1≤
≤
for
→−2 4
, so lim [[ ]] = lim ( − 1) =
→
⇔
54. (a) See the graph of
Since −1 ≤ cos on [−
−
on [−
= .
→ +
is not an integer.
= cos . 0 on [−
−
2), we have
= ( ) = [cos ]] = −1
2).
Since 0 ≤ cos 2 0) ∪ (0
− 1.
→ −
+ 1, so lim [[ ]] = lim → +
(c) lim [[ ]] exists
(−3) = −3.
lim
→ −
(ii) [[ ]] =
1 on [− 2].
1 2
= 4 − 1 = 3 and
= 7.
=1
( ) = lim (2 −
→1+
Now 3 =
( ) = lim− 4 −
2 0) ∪ (0
2], we have ( ) = 0
( ) = 1, we have lim ( ) = 1. →1
Since −1 ≤ cos
0 on (
2
], we have ( ) = −1 on (
2
].
Note that (0) = 1.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.3
CALCULATING LIMITS USING THE LIMIT LAWS
(b) (i) lim
→0
→0
→0−
(ii) As
→ (
2)− , ( ) → 0, so
(iii) As
→ (
2)+ , ( ) → −1, so
lim
( ) = 0.
2)−
→(
( ) = −1.
lim →(
2)+
(iv) Since the answers in parts (ii) and (iii) are not equal, lim →
in the open interval ( −
(c) lim ( ) exists for all →
2
( ) does not exist.
= − 2 and
) except
=
2.
]] + [− ]] is the same as the graph of ( ) = −1 with holes at each integer, since ( ) = 0 for any
55. The graph of ( ) =
( ) = −1 and lim
integer . Thus, lim
→2−
( ) = − 1, so lim ( ) = −1. However, →2
→2+
(2) = [[2 ] + [[− 2]] = 2 + ( −2) = 0, so lim ( ) = (2). →2
56. lim
1−
0
→ −
2
=
2
0
√ 1 − 1 = 0. As the velocity approaches the speed of light, the length approaches 0.
is not defined for
A left-hand limit is necessary since 57. Since ( ) is a polynomial, ( ) =
lim ( ) = lim →
0
→
=
+
0
1
+
0
+
+
1
+
2
2
+
1
2
2
2
2
.
+ ··· +
+ ··· +
+ ··· +
. Thus, by the Limit Laws, =
+
0
1
lim →
+
2
lim →
2
+ ··· +
lim →
= ( )
Thus, for any polynomial , lim ( ) = ( ). →
58. Let ( ) =
( ) where ( ) and ( ) are any polynomials, and suppose that ( ) = 0. Then ( )
lim ( ) = lim →
→
( )
lim ( ) →
=
( )
[Limit Law 5]
lim
( )
= ( ).
( )− 8 · ( − 1) = lim
−1
→1
[Exercise 57]
( )
( )− 8 →1
=
( )
→
59. lim [ ( ) − 8] = lim
· lim ( − 1) = 10 · 0 = 0. −1
→1
→1
Thus, lim ( ) = lim {[ ( ) − 8] + 8} = lim [ ( ) − 8] + lim 8 = 0 + 8 = 8. →1
→1
Note: The value of lim
→1
lim
( )− 8
→1
→1
( )− 8 does not affect the answer since it’s multiplied by 0. What’s important is that −1
exists.
−1
→1
( )
( ) 60. (a) lim
( ) = lim →0
→0
(b) lim
¤
( ) = 0 and lim+ ( ) = 0, so lim ( ) = 0.
( )
= lim
· 2
( )
2
= lim →0
= lim
· lim 2
( )
2
=5·0 =0
→0
lim
= 5 0 = 0
89
→0
→0
61. Observe that 0 ≤
( )≤
·
2
2
→0
2
·
→0
·
for all , and lim 0 = 0 = lim →0
→0
2
. So, by the Squeeze Theorem, lim ( ) = 0. →0
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90
¤
CHAPTER 2
LIMITS AND DERIVATIVES
62. Let ( ) = [ ]] and ( ) = −[[ ]]. Then lim
( ) and lim ( ) do not exist [Example 10]
→3
→3
but lim [ ( ) + ( )] = lim ([[ ]] − [[ ]]) = lim 0 = 0. →3
→3
( ) and ( ) = 1 −
63. Let ( ) =
Thus, either
→0
−2
3−
−1
→2
−2
√ 6− ·√
+2
√ 3− ·√
+1
→2
3−
−1
6−
+2
3−
+1
− 22
→2
2
3−
(2 − )
√ 3−
(2 − )
6−
65. Since the denominator approaches 0 as
+1
= lim
→2
6−
− 12
√ 3−
+1
+2
= lim √ →2 6−
+2
2
→−2
+ 15 + 18 2
+
= lim
−2
→−2
⇔
2
12 − 2 + + 3 = 0
3( + 2)( + 3)
−
2
+1
3−
√ ·√ −1 6−
+2
1
2
= 1 − ( − 1)2
⇔
shrinking circle to find the -coordinate, we get
2
1
+ +3 = 0
⇔
= 15. With
+ 3)
=
−1
⇔
= 15, the limit becomes
3(−2 + 3) −2 − 1
−
1− =
1 2 2
1 2 4
−0
−
( − 0). We set
are
= −1.
−3
2
, and take the limit as +
2
= 1 +2 − 1
2 2
+
2
=
2
and ( − 1)2 +
2
=
→ 0. The coordinates o
⇔ ⇔
= 2
=
1 2 . 2 2
2
= 1. Eliminating
1−
1
2
⇔
1−
1 2 4
from these
. The equation of the line joining
= 0 in order to find the -intercept, and get
−
=
4 1 2 2
are (0 ).
Substituting back into the equation of the
2
(the positive -value). So the coordinates o
3
=
as functions of . Then we can find the equation of the line determined
and
is the point of intersection of the two circles 2
3−
+
→−2
by these two points, and thus find the -intercept (the point
equations, we get
−4
=
3
= lim
( − 1)( + 2)
66. Solution 1: First, we find the coordinates of
The point
6−
→ −2, the limit will exist only if the numerator also approaches
lim 3 → − 2. In order for this to happen, we need →−2
3(−2)2 + (−2) + + 3 = 0
→0
+2
+1
√
= lim →2
3− √ ·√
2
√
3
→0
√ 6− √
= lim
lim
→0
= lim
√ 6−
0 as
is the Heaviside function defined in Exercise 1.3.59.
( ), where
or is 0 for any value of . Then lim ( ) and lim ( ) do not exist, but lim [ ( ) ( )] = lim 0 = 0.
√ 6−
64. lim √
→3
1 and
1 2 4
is thus
1
=−
2
1−
Now we take the limit as
→ 0+ : lim →0+
So the limiting position o
− 21
2
1−
2
1 2 4
+1 =2
= 1 2 4
1−
−1
= lim 2 →0+
1 4
1−
1 4
2
2
1−
1 2
+1
4
−1
+ 1 = lim 2
√ 1 + 1 = 4.
→0+
is the point (4 0).
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.4
THE PRECISE DEFINITION OF A LIMIT
¤
91
Solution 2: We add a few lines to the diagram, as shown. Note that = 90◦ (subtended by diameter
∠
). It follows that ∠
(subtended by diameter ◦
∠
= 90 − ∠
4
, implying that
= 90◦ = ∠
). So ∠
=∠
=∠
. Since 4
=
. As the circle
plainly approaches the origin, so the point
. Also
is isosceles, so is 2
shrinks, the point
must approach a point twice
as far from the origin as , that is, the point (4 0), as above.
2.4
The Precise Definition of a Limit
1. If ( ) − 1
0 2, then −0 2
true if 0 7
( )− 1
1 1, so we can choose
⇒
02
08
( )
1 2. From the graph, we see that the last inequality is
= min {1 − 0 7 1 1 − 1} = min {0 3 0 1} = 0 1 (or any smaller positive
number). 2. If ( ) − 2
0 5, then −0 5
true if 2 6 Note that
( )− 2
⇒
05
15
( )
2 5. From the graph, we see that the last inequality is
= min {3 − 2 6 3 8 − 3} = min {0 4 0 8} = 0 4 (or any smaller positive number).
3 8, so we can take = 3.
3. The leftmost question mark is the solution of
√
2 42 = 5 76. On the left side, we need − 4
= 1 6 and the rightmost,
√
= 2 4. So the values are 1 62 = 2 56 and
2 56 − 4 = 1 44. On the right side, we need − 4
To satisfy both conditions, we need the more restrictive condition to hold — namely, − 4
5 76 − 4 = 1 76.
1 44. Thus, we can choose
= 1 44, or any smaller positive number. 4. The leftmost question mark is the positive solution of
solution of
2
= 3 , that is,
=
2
the right side, we need − 1
3
2
= 12, that is,
1 √
=
. On the left side, we need − 1
1 √ 2
2 3 2
2,
and the rightmost question mark is the positive − 1 ≈ 0 292 (rounding down to be safe). On
− 1 ≈ 0 224. The more restrictive of these two conditions must apply, so we choose
= 0 224 (or any smaller positive number). 5.
From the graph, we find that 4
−
1
≈ 0 675
≈ 0 876, so
when
Thus, we choose the smaller of 6.
⇒
1
1
+
4
≈ 2
= tan 4
= 0 8 when
≈ 0 675, so
− 0 675 ≈ 0 1106. Also,
≈ 0 876
⇒
2
= 0 876 −
= tan 4
=12
≈ 0 0906.
= 0 0906 (or any smaller positive number) since this is and
2.
From the graph, we find that = 2 ( 2 + 4) = 0 3 when = 23, so 1 1 − 1 = 23 ⇒ = 2 ( 2 + 4) = 0 5 when = 2, so 1 = 3 . Also,
1+
2
= 2
⇒
2
= 1. Thus, we choose =
number) since this is the smaller of
1
and
1 3
(or any smaller positive
2.
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92
¤
CHAPTER 2
LIMITS AND DERIVATIVES
= 0 2, we find that
From the graph with
7.
≈ 1 9774, so 2 − 3
= 2
⇒
≈ 1 9774
1
− 3 + 4 = 6 2 when
≈ 2 0219
2
⇒
= 0 0219 (or any smaller positive
number) since this is the smaller of 1
− 3 + 4 = 5 8 when
≈ 0 0226. Also,
1
≈ 2 022, so 2 +
≈ 0 0219. Thus, we choose
For = 0 1, we get
3
=
1
and
≈ 0 0112 and
2.
≈ 0 0110, so we choose
2
= 0 011 (or any smaller positive number). From the graph with
8.
= 0 5, we find that
=(
≈ −0 303, so
1
≈ 0 215, so
≈ 0 215. Thus, we choose
2
≈ 0 303. Also,
− 1)
2
=(
positive number) since this is the smaller of For = 0 1, we get
1
≈ 0 052 and
2
− 1)
2
= 1 5 when
= 2 5 when
= 0 215 (or any smaller 1
and
2.
≈ 0 048, so we choose
= 0 048 (or any smaller positive number). 9. (a)
The first graph of
=
1 shows a vertical asymptote at ln( − 1)
≈ 2 01 (more accurately, 2 01005). Thus, we choose (b) From part (a), we see that as 1
lim →2+
ln( − 1)
10. We graph
= csc2
csc2
11. (a)
and
2
and
(b) − 1000 ≤ 5 995
≤
≤
= 0 01 (or any smaller positive number).
gets closer to 2 from the right,
increases without bound. In symbols,
≈ 0 044. Thus, if 0
1005
−
= 1000, we get
⇒
= 1000 cm2 ⇒
≈ 3 186, so
= 500. The graphs intersect at
500. Similarly, for
=
= 100 when
= ∞.
= 3 186 −
we choose
= 2. The second graph shows that
−5 ≤ ⇒
2
2
= 3 173 −
= 1000
− 1000 ≤ 5
17 7966 ≤
0 044, then
⇒ ⇒
≤ 17 8858.
≈ 0 031.
2
=
1000
1000 − 5 ≤ 1000
−
⇒
=
2
≤ 1000 + 5
995
1000
(
0)
≈ 17 8412 cm.
⇒
≈ 0 04466 and
1005
−
1000
≈ 0 04455. So
if the machinist gets the radius within 0 0445 cm of 17 8412, the area will be within 5 cm2 of 1000.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.4
(c)
is the radius, ( ) is the area,
¤
THE PRECISE DEFINITION OF A LIMIT
93
is the target area (1000 cm2 ), is the magnitude
is the target radius given in part (a),
of the error tolerance in the area (5 cm2 ), and is the tolerance in the radius given in part (b). 12. (a)
=01 01
2
2
+ 2 155
+ 20 and
+ 2 155 + 20 = 200
(b) From the graph, 199 ≤ (c)
⇒
[by the quadratic formula or
≈ 33 0 watts (
from the graph
⇒
= 200
≤ 201
0) ⇒
32 89
is the input power, ( ) is the temperature,
33 11. is the target input power given in part (a),
is the target temperature (200),
is the tolerance in the temperature (1), and is the tolerance in the power input in watts indicated in part (b) (0 11 watts). 01 01 , so = = 0 025. 13. (a) 4 − 8 = 4 − 2 0 1 ⇔ − 2 4 4 0 01 (b) 4 − 8 = 4 − 2
0 01
⇔
− 2
0 01 , so
4
=
= 0 0025.
4
14. (5 − 7) − 3 = 5 − 10 = 5( − 2) = 5 − 2 . We must have ( ) −
− 2
5. Thus, choose
15. Given
0, we need
=
5. For
0 such that if 0
= 0 1,
= 0 02; for
− 3
= 0 05,

, so 5 − 2
= 0 01; for
= 0 01,
⇔ = 0 002.
, then 1
(1 +
1
)− 2
. But (1 +
3 1 3
0
1
)− 2
⇔
−1
3
− 3
⇔
− 3
− 3
⇒
(1 +
⇔
3
3 . So if we choose = 3 , then 1 3
)− 2
. Thus, lim (1 + →3
1 3
) = 2 by
the definition of a limit.
16. Given
0, we need
(2 − 5) − 3 2 − 4 0
− 4
. But (2 − 5) − 3 ⇔
− 4
0 such that if 0
− 4
⇒
⇔
, then
2 − 8
2. So if we choose
(2 − 5) − 3
=
⇔ 2, then
. Thus, lim (2 − 5) = 3 by the →4
definition of a limit.
17. Given
0, we need
(1 − 4 ) − 13 −4 − 12 we choose
− (−3)
. But (1 − 4 ) − 13 ⇔
=
0 such that if 0
−4 + 3
4, then 0
− (−3)
, then
⇔ ⇔
− (−3) ⇒
4. So if
(1 − 4 ) − 13
.
Thus, lim (1 − 4 ) = 13 by the definition of a limit. →−3 x
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
94
¤
CHAPTER 2
18. Given
LIMITS AND DERIVATIVES
0, we need
(3 + 5) − (−1)
=
. But (3 + 5) − (−1)
⇔
3 + 6
⇔
3 + 2
3, then 0
− (−2)
0 such that if 0
⇒
+ 2
, then
⇔
+ 2
3. So if we choose
(3 + 5) − (−1)
. Thus,
lim (3 + 5) = −1 by the definition of a limit. →−2
2+ 4 19. Given
0, we need
− 1
0 such that if 0
2+ 4 −2
, then
−2
. But
3 4 −4 ⇔
3
2+ 4 3
3
. Thus, lim
→1
0, we need ⇔
4
4 5
−
− (−5)
=
4
, then 0
− 1
⇒
= 2 by the definition of a limit.
3
4
− 10
− 10 ⇔
, then 3 −
→10
0, we need
− 4
− 2 −8
22. Given
5
− 10
=
6
⇔
⇒
− 4
, then 0
− (−5)
− 10
⇔ ⇒
4
2
− 4
, then
+ 2 − 6
= 4] ⇔
⇒
+ 2 − 6
−2 −8 −6 −4
− 4
⇔
. So choose = . Then
. By the definition of a limit, lim →4
(3 + 2 )(3 − 2 )
0 such that if 0
6
⇔
+ 1 5
3 − 2 − 6
( − 4)( + 2) −6 −4
⇒
2
6 −
3+ 2
5
. So if we choose
4 5
) = −5 by the definition of a limit.
0 such that if 0
0, we need
+ 1 5
4 5
. But 3 −
−
−4
−4
− (−5)
4 5
4
. Thus, lim (3 −
( − 4)( + 2)
2
. So if we choose
4
− 1
5
21. Given
0
⇔
0 such that if 0
5
3−
− 1
3
2+ 4 −2
20. Given
8−
3
4
⇔
3
−2 −8
= 4] ⇒
= 6.
−4 , then
9− 4 3+ 2
[ = −1 5]
⇔
2
−6
⇔
−2 − 3
⇔
⇔
−2 + 1 5
− 2. So choose
=
2. Then 0
+ 1 5
⇒
+ 1 5
2
⇒
−2 + 1 5
⇒
−2 − 3
⇒
3 − 2 − 6
⇒
(3 + 2 )(3 − 2 )
−6
= −1 5]
⇒
3+ 2 By the definition of a limit,
lim →−1 5
23. Given
0, we need
9− 4
9− 4
2
−6
3+ 2
2
= 6.
3+ 2
0 such that if 0
−
, then −
. So
= will work.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
.
SECTION 2.4
24. Given
0, we need
0 such that if 0
−
, then −
0 such that if 0
− 0
, then
¤
THE PRECISE DEFINITION OF A LIMIT
95
. But − = 0, so this will be true no matter
what we pick. 25. Given
0, we need − 0
Then 0 26. Given
⇒
0, we need
−0
. Thus, lim
⇒
3
2
→0
− 0
0 such that if 0
− 0
Then 0
2
−0
3
−0
2
⇔
0, we need
3
= . Thus, lim
−0
3
, then
→0
− 0
0 such that if 0
√

. Take
=
√
.
= 0 by the definition of a limit. ⇔
⇔
3
√ 3

. Take
=
√ 3
= 0 by the definition of a limit.
27. Given
⇔
2
− 0
, then
. But
= . So this is true if we pick
= .
Thus, lim = 0 by the definition of a limit. →0
28. Given
0, we need
√ 8 6+ √ 8 6+
⇔ −0
− (−6)
0 such that if 0
. Thus,
−0
√ 8 6+
. But
8 ⇔ − (−6) . So if we choose = 8 , then 0 √ 8 6 + = 0 by the definition of a right-hand limit.
8
6+
√ 8 6+
, then
lim
−0
⇔
− (−6)
⇒
→−6+
29. Given
0, we need
( − 2)2 2
lim →2
. So take
− 2
, then
− 2
⇔
0, we need
+2 − 8
+ 4
⇔
2
→2
31. Given
−4
− 2
. But (
⇔
. Thus,
. Thus our goal is to make − 2 small enough so that its product with + 4
⇒
7, so (
2
If − 2
2
, then (
1. Then −1 − 2
−2
1
7. Choose
⇒
1
2
+ 2 − 7) − 1
⇒
3
= min {1
5
⇔
+4
⇒
7
− 2
7}. Then if 0
+ 2 − 7) − 1 = ( + 4)( − 2) = + 4 − 2
, we
7( 7) = , as
+ 2 − 7) = 1 by the definition of a limit.
0, we need
0 such that if 0
whenever 0 5. So take
+ 2
− (−2)
= min { 5 1}. Then 0
0, we need
0 such that if 0
+ 2
1, that is, 1
− 8 = − 2
2
− 8 = − 2
2
3, then
2
− 2 +2 +4
2
, then
. Notice that if + 2
− 1 − 3 = ( + 2)( − 2) = + 2 − 2
32. Given
3
+ 2 − 7) − 1
− 4 +4
+ 4 − 2
7 and + 4
desired. Thus, lim (
3
( − 2)2
2
− 2
7, and this gives us 7 − 2
have − 2
2
⇔
0 such that if 0
is less than . Suppose we first require that − 2
2
− 4 +5 − 1 √ − 2 ⇔ 2
− 4 + 5 = 1 by the definition of a limit.
30. Given 2
0 such that if 0 √ = . Then 0
−1 −3
or upon simplifying we need
1, then −1
+2
⇒
5 and + 2
− 2
1
⇒
−5
−2
→−2
3
−8
. Now
3
− 8 = ( − 2)
2
19 − 2 . So if we take
+2 +4
19
= min 1
19
− 2
, then 0
· 19 = . Thus, by the definition of a limit, lim
→2
3
= 8.
2
− 1) = 3.
+2 +4 .
32 + 2(3) + 4 = 19 and so
+2 +4
⇒
5, so
( 5)(5) = . Thus, by the definition of a limit, lim (
, then
−3
⇒
.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
96
¤
CHAPTER 2
33. Given
LIMITS AND DERIVATIVES
0, we let
4
+3
= min 2
⇒
8
=
√ 9+
value of {
− 3
. If 0
8. Also − 3
+ 3
34. From the figure, our choices for 2
8
are
= 3−
1
2 };
1
1
that is,
≈ 0 891 and
smaller of 1 −
(b) Solving
8,
2
so
√ 9−
2
⇒
−2
− 9 = + 3 − 3
−3 8·
8
⇒
2
2
= . Thus, lim
→3
= 9.
and
− 3. The largest possible choice for is the minimum = min{
2}
1
=
3
1
and
=
2
35. (a) The points of intersection in the graph are (
with
, then − 3
1
√ 9+
− 3.
2 6) and (
2
3 4)
≈ 1 093. Thus, we can take to be the
2 2
− 1. So
=
2
− 1 ≈ 0 093.
+
+ 1 = 3 + gives us two nonreal complex roots and one real root, which is √ 2 3 216 + 108 + 12 336 + 324 + 81 2 − 12 . Thus, = ( ) − 1. ( )= √ 1 3 6 216 + 108 + 12 336 + 324 + 81 2
(c) If = 0 4, then ( ) ≈ 1 093 272 342 and 36. 1. Guessing a value for
Let 1
− 2
0
. But
−
0 such that
1
−
1 2
whenever
− 2
= 2
− 2
0 be given. We have to find a number 2−
1
= ( ) − 1 ≈ 0 093, which agrees with our answer in part (a).
1 . We find a positive constan
=
− 2 and we can make
− 2
⇒
such that
2
2
2
by taking − 2
= . We restrict
to lie in the interval
2 − 2 choose
1
⇒
1
1
1 3
⇒
1 6
1 2
1 2
1 2
⇒
Given
1 (as in part 1). Also − 2 2
0 we let 2 , so
= min {1 2 }. If 0 1
1 − 2 − 2 = 2
− 2
37. 1. Guessing a value for

√ −
=
1 is suitable. Thus, we should 2
−
Given
0, we must find
, then − 2
1
⇒
1 · 2 = . This shows that lim (1 →2 2 √
√
1 . So 2
= min {1 2 }.
2. Showing that works 1 2
3 so 1
0 such that
1
3
) = 12 .
√ −

whenever 0 √
− √
. But
⇒
= √ √ + (from the hint). Now if we
+
then
can find a positive constan
such that
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.4
− √
+
, and we take −
centered at . If − 1
+
√
1
, then − 1
2
2
. We can find this number by restricting
−
⇒
1 2
3
2
2
is a suitable choice for the constant. So −
1
1 2
1 2
2. Showing that
−
−
√
works
0, we let
√
+
1 2
− = √ √ +
38. Suppose that lim 1
1 2
√
1
2
2
→0
with 0
√
= . Therefore, lim
√
√
+
, and so
. This suggests that we let
√
1 2.
( )=
1 2.
This contradicts
1 + 2 √
√
=
0 such that 0 +
. Then ( ) = 0, so 0 −

1 2.

, then
, so
by the definition of a limit.
1

≤
1 2.
s
Hence, 1 −
1 2,
. For −
0,
⇔
1 2

( ) = 0,
( ) does not exist.
0 such that 0 1 2,
( )−
2
→0

⇒

⇒
1
Therefore, lim
= 12 , there exists
. Given
−
. If 0
→
( ) = 1, so 1
. Then ( ) = 1, so 1 −

+
2
⇒
0
39. Suppose that lim
number
1 2
(as in part 1). Also −
= 12 , there exists
. Given
1
2
−
0
= min
√
+ 1 . For 0
()
2
√
+
2+
()=
→0
so
√
. Given
√
⇒
1 2
√
+
2+
−
+
to lie in some interval
2
= min
√
+
√
⇒
1
2

so
1 2.
⇒
( )−
1 2.

Take any rational
Now take any irrational number with 1 2,
This contradicts
so lim ( ) does not →0
exist. 40. First suppose that lim →
−
Then
−
0
( )=
. Then, given
⇒
−
⇒

( )=
→ −
1
0
so ( ) −
Now suppose lim −
( )−
0 there exists so ( ) −
( )=
. Hence, lim
→ +
= lim+ ( ). Let
( )−

. Let be the smaller of
( )−

. Hence, lim ( ) =
. Since lim
→ +
1
and
2.
1 ( + 3)4
10,000
⇔
( + 3)4
. Thus, lim− ( ) =
( )=
, there exists −
2
⇒
10,000

. ⇒
+
+ 3
1 √ 4 10,000
( )=
, there exists
1
0 so that
+
2
− ⇔
→
⇔
( )−
.
→ −
Then 0
⇒
. Also
→
. So we have proved that lim ( ) =
1
−
0 so that 0
0 be given. Since lim
→


→
41.
97
−
√
=
¤
THE PRECISE DEFINITION OF A LIMIT
⇒
or
1
lim
+
( )=
→ −
⇔
0 so that
− (−3)
= lim → +
1 10
2
so
( ).
42. Given
0, we need 1
( + 3)4
lim →−3
1 ( +
⇔
0 such that 0
+ 3
+ 3
1 √ . So take 4
4 ⇒ 1 ( + 3)
1 . Then 0 = √ 4
+ 3
. Now
1 ( + 3)4
1 = √ 4
⇒
⇔ 1 ( + 3)4
= ∞. 3)4
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
, so
98
¤
CHAPTER 2
43. Given
LIMITS AND DERIVATIVES
0 we need
suggests that we take
44. (a) Let
0 so that ln =
whenever 0
. If 0
, then ln
ln
be given. Since lim ( ) = ∞, there exists
→
smaller of
1
and
2.
−
0 such that 0
2
−
Then 0
⇒
whenever 0
. This = −∞.
→0
⇒
2
ln
=
. By the definition of a limit, lim+ ln
=
−
0 such that 0
1
→
lim ( ) = , there exists
; that is,
( )−
⇒
1
1
⇒
− 1. Let be the
( )
+ 1 − ) + ( − 1) =
( )+ ( )
+ 1 − . Since
( )
. Thus,
lim [ ( ) + ( )] = ∞. → (b) Let
0 be given. Since lim ( ) =
( )− ( )
2
(c) Let
( )
2.5
2 . Let
⇒
→
= min {
1
2 }.
0, there exists
→
2 2
− 2
⇒
( )
. (Note that ⇒
→
2
⇒
0 ·
2
=
⇒
1
2 →
2 −
0 such that 0
0.) Let
, so lim
2
( ) ( )
−
1
2 }.
⇒
2
, so lim ( ) ( ) = ∞.
=
→
1
⇒ −
0 such that 0
2
= min {
⇒
1
0 such that 0
2
( ) = ∞, there exists
2. Since lim
0 and
( ) ( )
−
Then 0
−
0 such that 0
1
2. Since lim ( ) = ∞, there exists
( )
0 be given. Since lim ( ) =
( )− ( )
0, there exists
→
Then 0
⇒
2
−
⇒
( ) ( ) = −∞.
Continuity
1. From Definition 1, lim
→4
2. The graph of 3. (a)
( ) = (4).
has no hole, jump, or vertical asymptote.
is discontinuous at −4 since (−4) is not defined and at −2, 2, and 4 since the limit does not exist (the left and right limits are not the same).
(b)
is continuous from the left at −2 since
lim
→−2−
( ) = (−2).
is continuous from the right at 2 and 4 since
lim+ ( ) = (2) and lim+ ( ) = (4). It is continuous from neither side at −4 since (−4) is undefined. →2
→4
4. From the graph of , we see that is continuous on the intervals [−3 −2), (−2 −1), (−1 0], (0 1), and (1 3]. 5. The graph of = ( ) must have a discontinuity at 6. The graph of = ( ) must have discontinuities
= 2 and must show that lim
→2+
( ) = (2).
at
= −1 and lim
→−1−
= 4. It must show that
( ) = (−1) and lim+ ( ) = (4). →4
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.5
7. The graph of
= ( ) must have a removable
discontinuity (a hole) at at
8. The graph of
= 3 and a jump discontinuity
at
CONTINUITY
¤
99
= ( ) must have a discontinuity
= −2 with
lim
→−2−
( ) = (−2) and
= 5. lim
( ) = (−2). It must also show that
→−2+
lim
→2−
( ) = (2) and lim+ ( ) = (2). →2
9. (a) The toll is $7 between 7:00 AM and 10:00 AM and between 4:00 PM and 7:00 PM .
(b) The function
has jump discontinuities at = 7, 10, 16, and 19. Their
significance to someone who uses the road is that, because of the sudden jumps in the toll, they may want to avoid the higher rates between = 7 and = 10 and between = 16 and = 19 if feasible. 10. (a) Continuous; at the location in question, the temperature changes smoothly as time passes, without any instantaneous jumps
from one temperature to another. (b) Continuous; the temperature at a specific time changes smoothly as the distance due west from New York City increases, without any instantaneous jumps. (c) Discontinuous; as the distance due west from New York City increases, the altitude above sea level may jump from one height to another without going through all of the intermediate values — at a cliff, for example. (d) Discontinuous; as the distance traveled increases, the cost of the ride jumps in small increments. (e) Discontinuous; when the lights are switched on (or off ), the current suddenly changes between 0 and some nonzero value, without passing through all of the intermediate values. This is debatable, though, depending on your definition of current. 4
11. lim
( ) = lim
→−1
+2
3 4
→−1
=
lim
+ 2 lim
→−1
3
1 + 2( 1)3
= −
→−1
4
−
= ( 3)4 = 81 = −
By the definition of continuity, is continuous at = −1. lim 2 + 5 lim lim( 2 + 5 ) 2 2 +5 2 + 5(2) 14 →2 →2 →2 12. lim ( ) = lim = = = = (2). = →2
→2
2 +1
lim(2 + 1) →2
By the definition of continuity,
2 lim + lim 1 →2
is continuous at
→2
= 2.
2(2) + 1
5
( 1). −
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
100
¤
CHAPTER 2
LIMITS AND DERIVATIVES
√
13. lim ( ) = lim 2 3 →1
→1
2
+ 1 = 2 lim
→1
√ 3
2
+1 =2
√ = 2 3(1)2 + 1 = 2 4 = 4 = (1) By the definition of continuity, is continuous at 14. lim
4
( ) = lim 3
→2
√ 3
−5 +
2
+4
4
√ 3
+ 1) = 2 3 lim
→1
− 5 lim
→2
= 3(2)4 − 5(2) +
2
→1
+
3
lim (
→2
→
Thus,
=
→1
2
+ 4)
22 + 4 = 48 − 10 + 2 = 40 = (2)
By the definition of continuity, is continuous at = 2. 4, we have √ lim ( ) = lim ( + − 4 ) = lim
is continuous at
+ lim 1
→2
15. For
So
2
= 1.
= 3 lim
→2
lim (3
→
=
+
=
+
→
√
→
−4
− lim 4
lim →
√
+ lim
[Limit Law 1] [8, 11, and 2]
→
−4
[8 and 7]
= ( ) for every in (4 ∞). Also, lim+ ( ) = 4 = (4), so →4
is continuous from the right at 4.
is continuous on [4 ∞). −2, we have
16. For
−1
lim ( ) = lim →
→
3 +6
lim →
=
=
lim ( − 1) →
[Limit Law 5]
lim (3 + 6) →
− lim 1
3 lim →
→
[2 1 and 3]
+ lim 6 →
−1 = 3 +6
[8 and 7]
is continuous at = for every in (−∞ −2); that is, is continuous on (−∞ −2). 1 ( )= is discontinuous at = −2 because (−2) is undefined. +2
Thus, 17.
18.
1 +2
( )= 1
if
= −2
if
= −2
Here (−2) = 1, but so lim
→−2
lim
→−2−
( ) = −∞ and
( ) does not exist and
lim
→−2+
( ) = ∞,
is discontinuous at −2.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.5
19.
2 lim
( )=
lim
( )=
→−1−
→−1+
if
≤ −1
if
−1
+3
( )=
CONTINUITY
¤
101
lim ( + 3) = −1 + 3 = 2 and
→−1−
lim 2 = 2−1 =
→−1+
1 2.
Since the left-hand and the
at − 1 are not equal, lim
right-hand limits of
( ) does not exist, and
→−1
is discontinuous at −1. − −1
2
20.
2
( )= 1
2
lim ( ) = lim →1
→1
( )=
0 1−
if
=1 ( − 1)
= lim
−1
→1
2
( + 1)( − 1)
if
0
if
=0
if
0
lim ( ) = 1, but (0) = 0 = 1, so 2 22.
−5 −3 −3
( )= 6
lim ( ) = lim →3
2
2
2
( )=
−
+1
1
,
2
if
=3
if =3 −5 −3 (2 + 1)( − 3) = lim = lim (2 + 1) = 7, −3
→3
but (3) = 6, so 23.
→1
=
is discontinuous at 0.
→0
2
= lim
is discontinous at 1
cos 21.
=1
−
2
but (1) = 1, so
if
−3
→3
→3
is discontinuous at 3.
−2
=
( − 2)( + 1)
−2
=
+ 1 for
= 2. Since lim ( ) = 2 + 1 = 3, define (2) = 3. Then
−2
is
→2
continuous at 2. 2 3
24.
( )= 2
−8
=
−4
( − 2)(
+ 2 + 4)
( − 2)( + 2)
2
=
+2 +4 +2
for
= 2. Since lim ( ) = →2
4+4 +4 2+2
Then 25.
is continuous at 2. 2 2− −1 is a rational function, so it is continuous on its domain, (−∞ ∞), by Theorem 5(b). ( )= 2 2
26.
= 3, define (2) = 3.
( )=
+1 +1
2
=
+1
is a rational function, so it is continuous on its domain,
2
2
−
−1
(2 + 1)( − 1)
−∞ − 12 ∪ − 12 1 ∪ (1 ∞), by Theorem 5(b). ° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
102
¤
27.
3
CHAPTER 2
−2 =0
LIMITS AND DERIVATIVES
⇒
3
=2 ⇒
=
√ 3
√ 3 2, so
( )= 3
continuous everywhere by Theorem 5(a) and
√ 3
−2
has domain
−2
√3 −∞
2 ∪
√ 3 2 ∞ . Now
3
− 2 is
− 2 is continuous everywhere by Theorems 5(a), 7, and 9. Thus,
is
continuous on its domain by part 5 of Theorem 4. sin
28. The domain of
()=
Theorems 7 and 9, of Theorem 4,
is (−∞ ∞) since the denominator is never 0 [cos
2 + cos
sin
and cos
≥ −1
are continuous on R. By part 1 of Theorem 4, 2 + cos
⇒
≥ 1]. By
2 + cos
is continuous on R and by part 5
is continuous on R.
29. By Theorem 5(a), the polynomial 1 + 2 is continuous on R. By Theorem 7, the inverse trigonometric function arcsin
is
continuous on its domain, [−1 1]. By Theorem 9, ( ) = arcsin(1 + 2 ) is continuous on its domain, which is { −1 ≤ 1 + 2 ≤ 1} = { −2 ≤ 2 ≤ 0} = { −1 ≤ 30. By Theorem 7, the trigonometric function tan
the composite function
√ 4−
2
or
1
1+
( )=
≤ −1, so
+1
=
is continuous on its domain,

=
2
+
. By Theorems 5(a), 7, and 9,
is continuous on its domain [−2 2]. By part 5 of Theorem 4,
continuous on its domain, (−2 −
31.
≤ 0} = [−1 0].
2) ∪ (−
2) ∪ (
2
is defined when
+1
has domain (−∞ −1] ∪ (0 ∞).
≥0
tan ( )= √ 4−
2
is
2 2). ⇒
+ 1 ≥ 0 and
0 or
+ 1 ≤ 0 and
0
⇒
0
is the composite of a root function and a rational function, so it is
continuous at every number in its domain by Theorems 7 and 9. − 2
32. By Theorems 7 and 9, the composite function
is continuous on R. By part 1 of Theorem 4, 1 +
− 2
is continuous on R.
By Theorem 7, the inverse trigonometric function tan−1 is continuous on its domain, R. By Theorem 9, the composite − 2
( ) = tan−1
function
is continuous on its domain, R.
1+ 33. The function
=
1
is discontinuous at
1
1+
left- and right-hand limits at
34. The function
where tan
2
= 0 are different.
= tan2 is discontinuous at = ln tan2
any integer. The function
is 0, that is, at
discontinuous at
=
2
,
= 0 because the
=
. So
=
2
+
, where
is
is also discontinuous = ln tan2
is
any integer.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.5
35. Because
√ 20 −
is continuous on R and √ 20 −
is continuous on its domain, −
2
√ is continuous on − 20 ≤ √ lim ( ) = (2) = 2 16 = 8. ( )=
20 ≤
≤
¤
103
√ 20, the product
√ 20. The number 2 is in that domain, so
≤
2
√
CONTINUITY
is continuous at 2, and
→2
36. Because
is continuous on R, sin
is continuous on R, and
( ) = sin( + sin ) is continuous on R, so lim 5− 1+
is continuous on R, the composite function
( ) = ( ) = sin( + sin ) = sin
→
37. The function ( ) = ln
+ sin
= 0.
2
is continuous throughout its domain because it is the composite of a logarithm function
and a rational function. For the domain of , we must have
5− 1+
2
0, so the numerator and denominator must have the
√ √ same sign, that is, the domain is (−∞ − 5 ] ∪ (−1 5 ]. The number 1 is in that domain, so lim ( ) = (1) = ln →1
5− 1
is continuous at 1, and
= ln 2.
1+1 √
38. The function ( ) = 3
2 −2 −4
is continuous throughout its domain because it is the composite of an exponential function,
a root function, and a polynomial. Its domain is
2
−2 −4≥ 0 =
2

− 2 +1 ≥ 5 =
− 1 ≥
=
( − 1)2 ≥ 5
= (−∞ 1 −
5 √
5 ] ∪ [1 + √
√
The number 4 is in that domain, so
( )=
39.
1−
2
ln
if
≤1
if
1
is continuous at 4, and lim ( ) = (4) = 3 →4
By Theorem 5, since ( ) equals the polynomial 1 −
2
on (−∞ 1],
At
= 1, lim
( ) = lim (1 −
→1−
2
) = 1 − 1 = 0 and lim 2
→1−
( )=
40.
sin
if
cos
if
is continuous at
( ) = lim ln
4 ∞),
= 32 = 9.
= ln 1 = 0. Thus, lim ( ) exists and →1
→1+
= 1. We conclude that
is continuous on (−∞ ∞).
4 ≥
4
By Theorem 7, the trigonometric functions are continuous. Since ( ) = sin (
16−8−4
is continuous on (1 ∞).
→1+
equals 0. Also, (1) = 1 − 12 = 0. Thus,
√
is continuous on (−∞ 1].
on (1 ∞),
By Theorem 7, since ( ) equals the logarithm function ln
5 ∞)
is continuous on (−∞
function is continuous at
4) ∪ (
4 Similarly,
4 ∞)
lim
→(
4)+
lim
→(
( )=
( )=
4)−
lim
→(
4)+
cos
on (−∞ lim
→(
=1
4)−
√
sin
4) and ( ) = cos on √ = sin 4 = 1 2 since the sine
2 by continuity of the cosine function
lim
( ) exists and equals 1
at
4. Thus,
so
is continuous on (−∞ ∞).
→(
4)
√ 2, which agrees with the value (
4). Therefore,
is continuous at
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4,
104
¤
CHAPTER 2
LIMITS AND DERIVATIVES
2
41.
−1
if
if − 1 ≤
( )= 1
1
≥1
if
is continuous on (−∞ −1), (−1 1), and (1 ∞), where it is a polynomial, a polynomial, and a rational function, respectively. 2 Now lim ( ) = lim = 1 and lim + ( ) = →−1−
so
→−1−
is discontinuous at −1. Since (−1) = −1,
lim
= 1 = (1), so
is continuous from the right at −1. Also, lim− ( ) = lim →1
→1−
= 1 and
is continuous at 1.
→1+
→1+
2 42.
1
( ) = lim
= −1,
lim
→−1+
→−1
3− √
( )=
≤1
if
≤4
if 1 if
4
is continuous on (−∞ 1), (1 4), and (4 ∞), where it is an exponential, a polynomial, and a root function, respectively. Now lim
→1−
( ) = lim 2 = 2 and lim+ ( ) = lim+ (3 − ) = 2. Since (1) = 2 we have continuity at 1. Also, →1−
→1
→1
( ) = lim (3 − ) = −1 = (4) and lim
lim →4−
→4−
( ) = lim
→4+
√
= 2, so
is discontinuous at 4, but it is continuous
→4+
from the left at 4. +2 43.
if
0
if 0 ≤
( )= 2−
if
≤1 1
is continuous on (−∞ 0) and (1 ∞) since on each of these intervals it is a polynomial; it is continuous on (0 1) since it is an exponential. = 1, so Now lim ( ) = lim ( + 2) = 2 and lim ( ) = lim →0−
→0−
→0+
continuous from the right at 0. Also lim
→1−
at 1. Since (1) = ,
→ −
is continuous at 2
45.
( )=
+2
→0+
( ) = lim
→1−
= and lim+ ( ) = lim+ (2 − ) = 1, so →1
→1
is continuous on its domain. We need to check for continuity at =
( ) = lim
→ −
is
is discontinuous
is continuous from the left at 1.
44. By Theorem 5, each piece of
lim
is discontinuous at 0. Since (0) = 1,
3
. Therefore, if
and lim 2
2
→ +
=
( ) = lim → +
is a continuous function of .
2
, so lim 2
→
=
.
( )=
. Since 2
( )=
, 2
3
−
if
≥2
is continuous on (−∞ 2) and (2 ∞). Now lim− ( ) = lim →2
→2−
2
+2
= 4 + 4 and
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.5 3
lim+ ( ) = lim+ →2
→2
−
= 8 − 2 . So
to be continuous on (−∞ ∞), 2 −4 if −2 46.
( )=
2
−
2 −
+
= 2:
lim
( ) = lim →2−
lim+ ( ) = lim ( →2
At
6 = 4
⇔
= 23. Thus, for
3
≥3
if
→2−
⇔
105
2
2
At
4 +4 = 8− 2
¤
= 23 .
if 2 ≤
+3
is continuous ⇔
CONTINUITY
→2+
−4
= lim
−2 2
−
( + 2)( − 2)
= lim ( + 2) = 2 + 2 = 4
−2
→2−
→2−
+ 3) = 4 − 2 + 3
We must have 4 − 2 + 3 = 4, or 4 − 2 = 1 (1). lim ( ) = lim− ( 2 − + 3) = 9 − 3 + 3
= 3:
→3
→3−
lim+ ( ) = lim+ (2 − →3
→3
+ )= 6−
We must have 9 − 3 + 3 = 6 −
+
+ , or 10
− 4 = 3 (2).
Now solve the system of equations by adding −2 times equation (1) to equation (2). −8 + 4 = −2 10 − 4 = 2 = So
= 12 . Substituting
1 2
in (1) gives us −2 = −1, so =
for
1 2
3 1 to be continuous on (−∞ ∞),
as well. Thus, for
= = 12 . 47. If
and
are continuous and (2) = 6, then lim [3 ( ) + ( ) ( )] = 36 →2
3 lim ( ) + lim ( ) · lim ( ) = 36 →2
48. (a) ( ) =
→2
1
→2
and ( ) =
1 2
⇒
, so ( ◦ )( ) = ( ( )) = (1
◦ is the set of numbers
(b) The domain of
3 (2) + (2) · 6 = 36
all nonzero reals). Thus, the domain is
2
⇒
) = 1 (1
⇒ 9 (2) = 36 2
)=
2
⇒
(2) = 4.
.
in the domain of (all nonzero reals) such that ( ) is in the domain of (also 1 = 0 and = 0 = { = 0} or (−∞ 0) ∪ (0 ∞). Since ◦ is 2
the composite of two rational functions, it is continuous throughout its domain; that is, everywhere except = 0. 4 −1 ( 2 + 1)( 2 − 1) ( + 1)( + 1)( − 1) = ( 2 + 1)( + 1) [or 3 + 2 + + 1] 49. (a) ( ) = = = 2 −1
−1
−1
= 1. The discontinuity is removable and ( ) = 3 + 2 + + 1 agrees with for = 1 and is continuous on R. 3 − 2−2 ( − 2)( + 1) ( 2 − − 2) (b) ( ) = = = ( + 1) [or 2 + ] for = 2. The discontinuity = for
−2
−2
−2
is removable and ( ) = 2 + agrees with for = 2 and is continuous on R. (c) lim ( ) = lim [[sin ]] = lim 0 = 0 and lim ( ) = lim [[sin ]] = lim (−1) = −1, so lim → −
→ −
→ −
→ +
→ +
→ +
→
( ) does not
exist. The discontinuity at
=
is a jump discontinuity.
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106
¤
CHAPTER 2
LIMITS AND DERIVATIVES
50.
does not satisfy the conclusion of the
does satisfy the conclusion of the
Intermediate Value Theorem.
51.
2
( )=
Intermediate Value Theorem.
is continuous on the interval [31 32], (31) ≈ 957, and (32) ≈ 1030. Since 957
+ 10 sin
1000
1030,
there is a number c in (31 32) such that ( ) = 1000 by the Intermediate Value Theorem. Note: There is also a number c in (−32 −31) such that ( ) = 1000 52. Suppose that (3)
6. By the Intermediate Value Theorem applied to the continuous function
the fact that (2) = 8
6 and (3)
6 implies that there is a number in (2 3) such that ( ) = 6. This contradicts the fact
that the only solutions of the equation ( ) = 6 are
= 1 and
= 4. Hence, our supposition that (3)
follows that (3) ≥ 6. But (3) = 6 because the only solutions of ( ) = 6 are 53.
4
( )=
+
on the closed interval [2 3],
− 3 is continuous on the interval [1 2]
= 1 and
6 was incorrect. It
= 4. Therefore, (3)
(1) = −1, and (2) = 15. Since −1
0
6.
15, there is a number
in (1 2) such that ( ) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation
4
+
− 3 = 0 in the
interval (1 2) √ √ is equivalent to the equation ln − + = 0. ( ) = ln − + is continuous on the √ √ interval [2 3], (2) = ln 2 − 2 + 2 ≈ 0 107, and (3) = ln 3 − 3 + 3 ≈ −0 169. Since (2) 0 (3), there is a
54. The equation ln
−
=
√
number in (2 3) such that ( ) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation √ √ ln − + = 0, or ln = − , in the interval (2 3). = 3 − 2 is equivalent to the equation
55. The equation
[0 1], (0) = −2, and (1) =
− 1 ≈ 1 72. Since −2
+ 2 − 3 = 0.
− 1, there is a number in (0 1) such that ( ) = 0 by the
0
+ 2 − 3 = 0, or
Intermediate Value Theorem. Thus, there is a root of the equation 56. The equation sin
interval [1 2]
2
=
−
is equivalent to the equation sin
+ 2 − 3 is continuous on the interval
( )=
−
2
+
= 0.
(1) = sin 1 ≈ 0 84, and (2) = sin 2 − 2 ≈ −1 09. Since sin 1
= 3−
( ) = sin 0
−
, in the interval (0 1). 2
+
sin 2 − 2, there is a number in
(1 2) such that ( ) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation sin sin
2
=
0
−
2
+
= 0, or
− , in the interval (1 2).
57. (a) ( ) = cos
1
is continuous on the
−
3
is continuous on the interval [0 1], (0) = 1
0, and (1) = cos 1 − 1 ≈ −0 46
0. Since
−0 46, there is a number in (0 1) such that ( ) = 0 by the Intermediate Value Theorem. Thus, there is a root
of the equation cos
−
3
= 0, or cos
=
3
, in the interval (0 1).
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SECTION 2.5
(b) (0 86) ≈ 0 016
0 and (0 87) ≈ −0 014
¤
CONTINUITY
107
0, so there is a root between 0 86 and 0 87, that is, in the interval
(0 86 0 87). − 3 + 2 is continuous on the interval [1 2], (1) = −1
58. (a) ( ) = ln
−1
0
0, and (2) = ln 2 + 1 ≈ 1 7
0. Since
1 7, there is a number in (1 2) such that ( ) = 0 by the Intermediate Value Theorem. Thus, there is a root of − 3 + 2 = 0, or ln
the equation ln
(b) (1 34) ≈ −0 03
= 3−
0 and (1 35) ≈ 0 0001
, in the interval (1 2). 0, so there is a root between 1 34 and 1 35 that is, in the
interval (1 34 1 35). 59. (a) Let ( ) = 100
−
−1
(100) = 100
100
− 0 01
2
Then (0) = 100
− 100 ≈ −63 2
0 and
0. So by the Intermediate
Value Theorem, there is a number in (0 100) such that ( ) = 0. −
This implies that 100
100
= 0 01 2 .
(b) Using the intersect feature of the graphing device, we find that the root of the equation is 60. (a) Let ( ) = arctan
(1) =
= 70 347, correct to three decimal places.
+
− 1. Then (0) = −1
0 and
0. So by the Intermediate Value Theorem, there is a
4
number in (0 1) such that ( ) = 0. This implies that arctan = 1 − . (b) Using the intersect feature of the graphing device, we find that the root of the equation is 61. Let ( ) = sin
3
= 0 520, correct to three decimal places.
. Then
is continuous on [1 2] since
is the composite of the sine function and the cubing function, both
of which are continuous on R. The zeros of the sine are at 3
pertinent cube roots are related by 1 =
√ 3
[call this value ]
1
3 2
2
8
2. [By observation, we might notice that
3 , and that the =
√ 3
and
2 are zeros of .]
Now (1) = sin 1 [1
3 2
, so we note that 0
] and then on [
0, ( ) = sin
3 2
= −1
0, and (2) = sin 8
2], we see there are numbers and
in (1
) and (
0. Applying the Intermediate Value Theorem on 2) such that ( ) = ( ) = 0. Thus,
has at
least two -intercepts in (1 2). 62. Let ( ) =
2
− 3+ 1
inspection, we see that
. Then 1 4
=
17 16
is continuous on (0 2] since 0, (1) = −1
0, and (2) =
3 2
0. Appling the Intermediate Value Theorem on
1
1 4
is a rational function whose domain is (0 ∞). By
1 and then on [1 2], we see there are numbers and
in
4
1 and (1 2) such that ( ) = ( ) = 0. Thus,
least two -intercepts in (0 2).
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has at
¤
108
CHAPTER 2
63. (⇒) If
LIMITS AND DERIVATIVES
is continuous at , then by Theorem 8 with ( ) =
lim ( + ) =
+
we have
lim ( + ) = ( ).
→0
→0
(⇐) Let
0. Since lim ( + ) = ( ), there exists
0 such that 0
→0
( + ) − ( )
−
. So if 0
Thus, lim ( ) = ( ) and so →0
cos
→0
=
+ cos sin ) = lim (sin
→0
+ lim cos
→0
→0
lim sin
→0
= (sin )(1) + (cos )(0) = sin
→0
65. As in the previous exercise, we must show that lim cos( + ) = cos
to prove that the cosine function is continuous.
→0
lim cos( + ) = lim (cos cos − sin sin ) = lim (cos cos ) − lim (sin →0
→0
=
66. (a) Since
)( ) = lim
→0
→0
( )=
→
lim ( ) = →
and are continuous at , lim →
( ) = lim
of Limits: lim →
( )=
− lim sin
lim cos
lim cos
→0
lim sin
→0
→0
sin )
= (cos )(1) − (sin )(0) = cos
→
→
67.
→0
is continuous at , lim ( ) = ( ). Thus, using the Constant Multiple Law of Limits, we have
lim ( (b) Since
.
cos ) + lim (cos sin )
→0
lim cos
lim sin
, then ( ) − ( ) = ( + ( − )) − ( )
is continuous at .
→
64. lim sin( + ) = lim (sin
⇒

→
0 if
is rational
1 if
is irrational
( )
( )=(
)( ). Therefore,
is continuous at .
( ) = ( ) and lim ( ) = ( ). Since ( ) = 0, we can use the Quotient Law →
lim ( ) →
=
( )
=
( )
( ). Thus,
=
is continuous at .
( )
lim ( ) →
is continuous nowhere. For, given any number
and any
0, the interval ( −
+ )
contains both infinitely many rational and infinitely many irrational numbers. Since ( ) = 0 or 1, there are infinitely many numbers
0 if ( )=
68.
−
with 0
if
and ( ) − ( ) = 1. Thus, lim ( ) = ( ). [In fact, lim ( ) does not even exist.] →
→
is rational is continuous at 0. To see why, note that − ≤ ( ) ≤ , so by the Squeeze Theorem
is irrational
lim ( ) = 0 = (0). But →0
is continuous nowhere else. For if
= 0 and
0, the interval ( −
+ ) contains both
infinitely many rational and infinitely many irrational numbers. Since ( ) = 0 or , there are infinitely many numbers −
0
and ( ) − ( )
2. Thus, lim ( ) = ( ). →
69. If there is such a number, it satisfies the equation
called ( ). Now (−2) = −5
3
+1 =
0, and (−1) = 1
⇔
3
−
+ 1 = 0. Let the left-hand side of this equation be
0. Note also that ( ) is a polynomial, and thus continuous. So by the
Intermediate Value Theorem, there is a number between −2 and −1 such that ( ) = 0, so that = 70.
3
+2
2
−1
+
3
+
−2
= 0
with
⇒
(
3
+
− 2) + (
3
+2
2
3
+ 1.
− 1) = 0. Let ( ) denote the left side of the last
equation. Since is continuous on [−1 1], (−1) = −4
0, and (1) = 2
0, there exists a in (−1 1) such that
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.6
¤
LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
109
( ) = 0 by the Intermediate Value Theorem. Note that the only root of either denominator that is in (−1 1) is √ √ (−1 + 5 ) 2 = , but ( ) = (3 5 − 9) 2 = 0. Thus, is not a root of either denominator, so ( ) = 0 ⇒ = is a root of the given equation. 71.
4
( )=
sin(1
) is continuous on (−∞ 0) ∪ (0 ∞) since it is the product of a polynomial and a composite of a
trigonometric function and a rational function. Now since −1 ≤ sin(1 lim (−
4
→0
) = 0 and lim
4
→0
= 0, the Squeeze Theorem gives us lim ( →0
continuous at 0 and, hence, on (−∞ ∞). ( ) = 0 and lim ( ) = 0, so lim
72. (a) lim+ →0
→0
→0−
0, lim →
( ) = lim →
=
=
( ). For
( ) = 0, which is
) ≤ 1, we have − 4
sin(1
4
→
4
sin(1
)≤
4
. Because
)) = 0, which equals (0). Thus,
(0), and hence
is continuous at
( ) = lim (− ) = − =
0, lim
≤
→
( ). Thus
=
is
if
= 0. For
is continuous at
= ; that is, continuous everywhere. (b) Assume that
is continuous on the interval . Then for
∈ , lim ( ) = lim ( ) = ( ) by Theorem 8. (If →
→
is
an endpoint of , use the appropriate one-sided limit.) So is continuous on . (c) No, the converse is false. For example, the function ( ) =
1
if
≥0
−1
if
0
is not continuous at
= 0, but ( ) = 1 is
continuous on R. 73. Define ( ) to be the monk’s distance from the monastery, as a function of time (in hours), on the first day, and define ( )
to be his distance from the monastery, as a function of time, on the second day. Le
be the distance from the monastery to
the top of the mountain. From the given information we know that (0) = 0, (12) = consider the function
− , which is clearly continuous. We calculate that ( − )(0) = −
So by the Intermediate Value Theorem, there must be some time ( 0 ) = ( 0 ). So at time
2.6
, (0) =
0
0
and (12) = 0. Now
and ( − )(12) =
between 0 and 12 such that ( − )( 0 ) = 0
. ⇔
after 7:00 AM , the monk will be at the same place on both days.
Limits at Infinity; Horizontal Asymptotes
1. (a) As
(b) As
becomes large, the values of ( ) approach 5. becomes large negative, the values of ( ) approach 3.
2. (a) The graph of a function can intersect a
The graph of a function can intersect a horizontal asymptote.
vertical asymptote in the sense that it can
It can even intersect its horizontal asymptote an infinite
meet but not cross it.
number of times.
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110
¤
CHAPTER 2
LIMITS AND DERIVATIVES
(b) The graph of a function can have 0, 1, or 2 horizontal asymptotes. Representative examples are shown.
No horizontal asymptote ( ) = −2 (b) lim
3. (a) lim →∞
→3
( )=2
(e) Vertical:
→2−
→1
= 1,
= 3; horizontal:
= −2,
( ) = −1
(b) lim
→∞
(d) lim
Two horizontal asymptotes (c) lim ( ) = ∞
→−∞
(d) lim ( ) = −∞ 4. (a) lim
One horizontal asymptote ( )= 2
=2 (c) lim ( ) = −∞
→−∞
( ) = −∞
→0
(e) lim+ ( ) = ∞
(f ) Vertical:
→2
horizontal: 5. lim
( ) = −∞,
→0
lim
→∞
8. lim
→∞
( ) = ∞,
→2
( ) = 5,
→−∞
lim
6. lim
( ) = −5
( ) = 3,
lim
→∞
( ) = −∞,
( ) = 0,
9. (0) = 3,
lim
→0−
lim
( ) = ∞,
lim+ ( ) = 2,
lim
( ) = −∞,
lim
→2−
→2+
is odd
lim
→−∞
( ) = 0,
(0) = 0
lim+ ( ) = ∞, →4
lim
( ) = 0,
lim
( ) = −∞
(0) = 0, lim
( ) = −∞,
→4−
lim
→∞
( )=3
=2
lim
( ) = ∞,
→∞
10. lim ( ) = −∞, →3
( ) = 4,
( ) = −∞,
= −1,
( ) = −∞,
→−∞
→0−
→0
→−∞
7. lim
= 2;
→2
→−2+
→−2−
lim
( ) = ∞,
lim
= 0,
lim+ ( ) = ∞, →0
( ) = 2, lim →∞
is even
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.6 2
11. If ( ) =
LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
¤
111
2 , then a calculator gives (0) = 0, (1) = 0 5, (2) = 1, (3) = 1 125, (4) = 1, (5) = 0 78125,
(6) = 0 5625, (7) = 0 3828125, (8) = 0 25, (9) = 0 158203125, (10) = 0 09765625, (20) ≈ 0 00038147, (50) ≈ 2 2204 × 10−12 , (100) ≈ 7 8886 × 10−27 . It appears that lim
2
→∞
12. (a) From a graph of ( ) = (1 − 2
2
= 0.
) in a window of [0 10,000] by [0 0 2], we estimate that lim
→∞
( ) = 0 14
(to two decimal places.) From the table, we estimate that lim
(b)
→∞
( ) 10,000 100,000 1,000,000
→∞
5
2
0 135 308 0 135 333 0 135 335
−7
2
2
13. lim
+
−3
− 7)
2
(2
= lim →∞
2
(5
− 3)
+
→∞
=
[Divide both the numerator and denominator by
2
2
(the highest power of [Limit Law 5]
2)
lim 2 − lim (7 →∞
=
→∞
[Limit Laws 1 and 2]
→∞
=
2 5
=
lim →∞
3
=
lim →∞
[Theorem 5]
3
9
+8 − 4
3− 5 +
3
→∞
lim (3 →∞
3
−5
3
2
2
2
lim 9 + lim (8 =
→∞ →∞
3)
→∞
3 lim (1 =
3
3)
]
) [Limit Law 5]
+ 1) 2
→∞
2
3
+1
) − lim (4
− lim (5
9 + 8 lim (1 →∞
[Divide by
→∞
lim (3
=
3
−4
−5
3
[Limit Law 11]
−4
2
9+ 8
lim (9 + 8 =
[Limit Laws 7 and 3]
2)
→∞
2 − 7(0) 5 + 0 + 3(0)
+8 −4
)
) − 3 lim (1
=
3− 5 +
2
→∞
5 + lim (1
2)
→∞
→∞
3
)
) − lim (3
2 − 7 lim (1
9
2
→∞
lim 5 + lim (1 →∞
that appears in the denominator)]
)
−3
lim (5 + 1 →∞
=
2
2
lim (2 − 7
14. lim
( ) = 0 1353 (to four decimal places.)
2)
[Limit Laws 1 and 2] →∞
→∞
− 5 lim (1
)
+ lim 1
) − 4 lim (1
√ 9 + 8(0) − 4(0) 3(0) − 5(0) + 1 9 =
3
→∞
3
) [Limit Laws 7 and 3]
2)
+1 1
=
9 = 3
→∞
[The orem 5] ° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
112
¤
15. lim →∞
CHAPTER 2
3 −2
LIMITS AND DERIVATIVES
(3 − 2)
= lim →∞
2 +1
3− 2
= lim →∞
(2 + 1)
lim 3 − 2 lim 1 →∞ →∞
=
2+1
lim 2 + lim 1 →∞
1−
16. lim →∞
= lim
−
3
(1 −
2
→∞
+1
3
(
2
−
→∞
lim 1 − lim 1 →∞
17.
−2
lim →−∞
18.
2
→−∞
2
√
2
+
→∞
2
(2 −
(2
2
2)
2
(2
21. lim →∞
=
− 4 + 5)
3
→∞
+1
+ 1)
( − 1)2 (
2
0+ 1
2
(2
+ 3 − 5)
3 2
2
(2
= lim →∞
2
=
= −1
1
− 1)2 (
[
2
2
−1
1 2
0− 1 =
→∞
−5
1 2
2+ 3
4
→∞
4
[(
2
= lim →∞
(1 − 2
+ )
2]
2
)
(2 + 0)
=
2 )(1
+1
]
2 ][( 2
+1
=4
(1 − 0 + 0)(1 + 0)
)
4 2
22. lim √ →∞
4
2
= lim √ 4
23. lim →∞
√ 1+ 4 2−
3
6
+1
4
1+1
= lim →∞
√ 1+ 4 (2 −
lim
6
6
(
4
[since
+ 1)
4
(1 + 4
6)
2
=
√
for
0]
1 = √ =1 1+0 lim
3
=
3)
1
→∞
2
1
= lim →∞
1
= lim
→∞
+1
2
3
+4
→∞
6
[since
lim (2 →∞
lim (1
3
− 1) 6)
+ lim 4
3
=
√
6
for
1 2
2 2
+ 1)
− 2 + 1)
2 2
(2 + 1
2
=−
2+ 0− 0
3 2
[(2
= lim
+ )]
=2
2−0+0
3
+5
1+0
0− 1
3 2
+ 1)
2
4+0−0
3
= lim
→∞
+ )
2− 4
3 2
= lim
2
=
−2
4+ 6
0 − 2(0)
=
lim 1 →−∞
→−∞
→−∞
−1
2
√
→−∞
lim 1 +
lim
3
3 2
1
= lim
→−∞
2
1+1
− 2)
2
+6
=
2
3
→−∞
3
− 2 lim 1
lim 1
2
−2
1
lim
2
−
+3 −5
3 2
=
(4
√ −
20. lim
2
)
=0
1−0+0
3
+ lim 1
3
+1
0−0
=
2
→∞
+ 1)
→−∞
→∞
2
2
= lim
(
= lim
2 −
(
−4 +5
+
19. lim
−2
2
+6
3
2
√
→∞
3
4
lim
→−∞
+1
2
→∞
( − 2)
= lim
2
− lim 1
3
→∞
2+ 0
−1
1− 1
→∞
3
3
=
→∞
3
1
= lim
+ 1)
lim 1 =
3
)
3 − 2(0)
=
0]
=0
=
→∞ 3)
lim (2
− lim 1
→∞
→∞
=
→∞
0− 1
→∞
√ 2 0+ 4 = −2 = = −1 −1 √ 1+ 4 24.
lim
→−∞
2−
3
√ 1+ 4
6
=
lim
→−∞
(2 −
lim − =
→−∞
√
3
6
1
→−∞
2 lim (1
=
3)
lim −
3
6
=
lim (2
→−∞
−
+4 = lim 1
→−∞
6)
√
6
→−∞
− 3)
(1 + 4 3
[since
− 1)
lim (1
6)
3
=−
6
for
+ lim 4
→−∞
→−∞
2(0) − 1
− 0+4 −2 = =2 −1 −1
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
0]
SECTION 2.6
√ 25. lim
+3
4 −1
→∞
√
2
= lim
→∞
+3
lim
2
lim (4 − 1
[since
)
→∞
¤
113
2
→∞
=
(4 − 1)
2)
( +3
LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
√
=
for
0]
2
lim
1
→∞
=
lim 4 − lim (1 →∞
+3
26. lim
2
=∞
27. lim
√ 9
2
)
+
1+ 3
→∞
4− 1
since 1 + 3 → ∞ and 4 − 1 √ 9
2
−3
2
→ 4 as
√ 9
−3
+
2
→∞
= lim √ →∞ 9
2
+
−9
2
+3
+
+
→−∞
2
+3 +2
2
9
= lim
→−∞
2
= lim √ →∞ 9
√ 4 4
= lim
→−∞
2
=
29. lim
→∞
√
2
+
−
√
2
+
+
+3
√ 4
√ 4
√ 4 √ 4
2
2
− (3 )2
+
+3
2
1 = √
1
=
9+ 3
+3
1 3+3
6
+3 − 2 +3 − 2
2
= lim →−∞
+3 −2 3
[since
−
=−
2
√ −2
4+ 3
3 3 √ =− 4 − 4+ 0− 2 2
= lim
+
−
→∞
= lim →∞
( √
2
+
2
+
2
+ 2 +
1+
+
√ √
2
+ 2 +
)−( 2 + ) = lim √ √ 2 + →∞ + −
= lim →∞
√ √
+
1
=
3
+3 − 2
2
2
·
9+1
= lim √ →−∞ 4
+3 − 2 3
2
9
1
− (2 )2
+3
√ 2
2
→∞
+3
+3 +2
= lim
→−∞
2
+
+
1
= lim
→∞
2
√
→∞
+3
2
= lim
√ 4
√ 9
+3 = lim
9 2
9
lim
+
→ ∞.
√
= lim
→∞
28.
√ 0+3 3 = 4 4
=
4− 0
= lim
(4 − 1)
→∞
→∞
→∞
=
)
√
) + lim 3
→∞
( +3
= lim
4 −1
→∞
lim (1
+3
1+
+
2
√
+
2
+
− ) ] √ 2 + +
− = √ = √ 1+0+ 1+0
− 2
√
2
for
0]
30. For
0,
√
2
√ +1
2
→ ∞, we have
= . So as
√
2
+ 1 → ∞, that is, lim
√
2
+ 1 = ∞.
→∞
−3
4
31. lim →∞
3
−
2
+
+2
= lim →∞
4
( (
−3 3
−
2
3
+ )
divide by the highest power of in the denominator
→∞
3
+ 2)
since the numerator increases without bound and the denominator approaches 1 as 32. lim ( →∞
−
+ 2 cos 3 ) does not exist. lim
→∞
−
−3
= lim
1− 1
2
+1
2
+2
3
=∞
→ ∞.
= 0, but lim (2 cos 3 ) does not exist because the values of 2 cos 3 →∞
oscillate between the values of −2 and 2 infinitely often, so the given limit does not exist.
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114
¤
33.
lim (
CHAPTER 2 2
LIMITS AND DERIVATIVES
7
+2
→−∞
→−∞
1
5
+ 2 → 2 as 2
Or: lim
1+
lim
7
→−∞
2
= lim
= −∞ because
[factor out the largest power of ]
(1 +
= lim
) 6
→−∞
+1
1+2
→−∞
6 4
+2
(
4
1
divide by the highest power
+
4
+ 1)
4
4
of
= lim
→−∞
in the denominator
→ ∞, → ∞. lim arctan(
. As
) = lim arctan = →∞
→∞
3
36. Divide numerator and denominator by
3
: lim →∞
37. lim →∞
1−
= lim →∞
1+2
38. Since 0 ≤ sin2
(1 −
)
= lim →∞
(1 + 2 )
≤ 1, we have 0 ≤
sin 2 2
sin
Theorem, lim →∞
2
39. Since −1 ≤ cos
lim
−2
→∞
3
−
−3
+
−3
−1
1
+2
1
≤ 2
+1
1+ 1
41. lim [ln(1 +
→ −∞.
1−
−6
1+
−6
=−
0+ 2
=
1−0
=1
1+0
1 2
1
. We know that lim 0 = 0 and lim →∞
→∞
+1
2
+1
+1 −2
−2
0, we have −
−1
→ 0+ , → −∞. lim+ tan →0
) − ln(1 + )] = lim ln
→∞
1+
→∞
≤ −2
−2
cos
≤
−2
. We know that lim (− →∞
(ln ) = lim tan−1 = − →−∞
2
= ln
42. lim [ln(2 + ) − ln(1 + )] = lim ln
lim →∞
1+
→∞
→∞
→0+
= 0 since
( ) = lim →0+
ln
2+ 1+
−2
) = 0 and
cos ) = 0.
1+
2
by (4).
1
2
= ln
lim →∞ 1
1+
+ +1
parentheses is ∞.
43. (a) (i) lim
= 0, so by the Squeeze
= 0.
= 0, so by the Squeeze Theorem, lim (
2
0− 1
=∞
2
≤ 1 and
As
=
→∞
2 4
by (3).
2
= lim
1
→∞
40. Let = ln
→ −∞ and
= −∞.
5
since the numerator increases without bound and the denominator approaches 1 as 35. Let =
7
5
→ −∞.
+2
→−∞
34.
1
7
) = lim
= lim ln →∞
→ 0+ and ln
2
+1
1
+1
→ −∞ as
= ln
1 1
→ 0+ .
= ln 1 = 0
= ∞, since the limit in
(ii) lim
→1−
→1−
(iii) lim
= −∞ since
( ) = lim
→1+
→ 1− .
ln = ∞ since
( ) = lim
→1+
→ 0− as
→ 1 and ln
→ 1 and ln
→ 0+ as
→ 1+ .
ln
(b)
(c) ( ) 10,000
1085 7
100,000
8685 9
1,000,000
72,382 4
It appears that lim →∞
( ) = ∞.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.6
44. (a) lim
1
−
→∞
→∞
since
2
( ) = lim
2
¤
(e)
=0
ln 1
→ 0 and
LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
→ 0 as
→ ∞.
ln 2
(b) lim+ ( ) = lim+ →0
since
→0
2
1
→ ∞ and
−
1 ln
=∞
→ 0 as
→ 0+ .
ln ( ) = lim
(c) lim
2
− 1
→1−
→1−
(d) lim+ ( ) = lim →1+
= ∞ since
2
ln 2
→1
−
1
→ 2 and
→ −∞ as
→ 1− .
1 ln → ∞ as
→ 1+ .
ln
1 ln
= −∞ since
2
→ 2 and
45. (a)
(b) ( )
From the graph of ( ) =
√
2
+
−10,000
−0 499 962 5
−100,000
−0 499 996 2
−1,000,000
−0 499 999 6
From the table, we estimate the limit to be −0 5.
+ 1 + , we
( ) to be −0 5.
estimate the value of lim
→−∞
√ √
(c) lim
→−∞
2
√ +
+1+
= lim
= lim →−∞
= Note that for 1√
2
+
0, we have
+1 =−
1 √ √
46. (a)
2
→−∞
2
√
2
+
+ √ 2 +
+1 +
( + 1)(1 √
2
2 2
+
)
+1 −
+1 − +1 −
→−∞
)
lim
→−∞
√
1 + (1
= lim (1
=
−
1 + (1
+ 2
+1 − +1 −
+
)
) + (1
2)
−1
1+0 1 √ =− 2 − 1+0+0− 1 2
+
= = − , so when we divide the radical by , with + 1 = − 1 + (1
) + (1
2 ).
0, we get
2
115
aces) the limit to be 1 4434. (b) ( )
From the graph of √ ( )= 3 2 +8 +6 − √ 3 2 + 3 + 1, we estimate
F r o
(to one decimal place) the value of lim ( ) to be 1 4.
m → ∞
t h e t a b l e , w e e s t i m a t e ( t o f o u r d e c i m a l p l
10,000
1 443 39
100,000
1 443 38
1,000,000
1 443 38
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116
¤
CHAPTER 2
(c) lim
LIMITS AND DERIVATIVES
√ 3
( ) = lim
→∞
→∞
+8 +6 − 3
2
3 = lim √ →∞ 3
5+ 4
→±∞
+8 +6+
3+8
→±∞
2
+
5
=
+ 3 → 0+ as
+3 +1
0+ 4
3
2
+8 +6+
√ 3
)
2
+ 3 + 1 (1 ) √ 5 5 5 3 = = = 1 443376 √ √ √ ≈ 6 3+ 3 2 3
2
1+3
= 4, so
1+ 0
, so
( ) = −∞
lim
+3 since 5 + 4 → −7 and
2
√
→∞
+1
+4
5+ 4
= ( )=
√ 3
(5 + 5)(1
+3 +1
3 +3
→±∞
= 4 is a horizontal asymptote.
+8 + 6 + 2 +3 +1
+3 +1
= lim
( + 3)
2
= lim 3
2
+6
(5 + 4 )
= lim
+3
2
5+5
→∞
lim
√ +3 +1 3 √ 2 +8 +6+ 3
2
√ 2
= lim
47.
√ 3 √ 3
+8 +6 −
2
→−3+
→ −3+ . Thus,
= −3 is a vertical
asymptote. The graph confirms our work. 48.
→±∞
2
2
lim
2
3
+1
=
+2 −1
→±∞
so
=
is a horizontal asymptote.
2
3
3 1 3
=
2
2
= ( )=
3
The denominator is zero when
2
=
−1
3+ 2
2
2
2+ 1
→±∞
2
+ 1)
+ 2 − 1)
2
(3
= lim
2
2
(2
lim
+1
2
=
+2 − 1
2
2
+1
.
(3 − 1)( + 1)
and −1, but the numerator is nonzero, so
=
= −1 are vertical asymptotes.
1 3 and
The graph confirms our work. 2 49.
lim →±∞
2
2
2
+
−1
+
−2
2
−1
+
→±∞
2
= lim −2
+
→±∞
1+
2
lim 2 + →±∞
lim
→±∞
= ( )=
2
2
2
lim →−2+
+
−1
+
−2
=
( ) = −∞, lim →1−
1
− lim
→±∞
= lim 1 +
1 −
2+
2
= lim
1
lim
→±∞
1
− 2 lim
→±∞
(2 − 1)( + 1) ( + 2)( − 1)
=
2
→±∞
→1+
1 2
1 lim
2 −
1+
2
2+ 0− 0
2
1
lim
=
1 + 0 − 2(0)
2
( ) = ∞,
→−2−
( ) = −∞, and lim
−
→±∞
2 −
1
1
→±∞
, so
1
2
2+
lim
( ) = ∞. Thus,
= −2
= 2,
so
= 2 is a horizontal asymptote.
and
= 1 are vertical asymptotes. The graph confirms our work. 4
1+ 50.
lim →±∞
1+ 2
−
4
=
→±∞
4
4
lim 2
−
= lim 4
→±∞
0+ 1 = −1, 0− 1
4
1 2
4
=
1
1
so
lim
+1 = −1
→±∞
lim
→±∞
4
1 +1
lim
= 1 2
−1
→±∞
lim
→±∞
4
1 2
+ lim 1 →±∞
− lim 1 →±∞
= −1 is a horizontal asymptote.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.6 4
1+ 2 −
= ( )=
4
1+ −
=
2 (1
4
4
2)
=
2 (1
1+ . The denominator is + )(1 − )
= 0, −1, and 1, but the numerator is nonzero, so
zero when
LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
= −1, and
= 0,
→ 0, the numerator and
= 1 are vertical asymptotes. Notice that as
denominator are both positive, so lim ( ) = ∞. The graph confirms our work. →0 −
3
51.
= ( )= 2
(
=
−6 +5
2
− 1)
( + 1)( − 1)
=
( − 1)( − 5)
The graph of is the same as the graph of at
( + 1)
( − 1)( − 5)
= ( ) for
= 1.
−5
with the exception of a hole in the 2
graph of
=
+
= 1. By long division, ( ) =
=
+6+
30
−5
.
−5
→ ±∞, ( ) → ±∞, so there is no horizontal asymptote. The denominator
As
of is zero when
( ) = −∞ and lim
= 5. lim
→5−
→5+
vertical asymptote. The graph confirms our work. 2 2 1 · 52. lim = lim = lim →∞
lim
−5 2
2(0)
= −5
0− 5
−5 =0
⇒
→−∞
when
−5
→∞
lim
2
→∞
1
= 0, so
= 5
1 − (5
2
= )
= 2, so
= 2 is a horizontal asymptote.
1− 0
= 0 is a horizontal asymptote. The denominator is zero (and the numerator isn’t)
⇒
= ln 5.
= ∞ since the numerator approaches 10 and the denominator
approaches 0 through positive values as
→(ln 5)−
2
= 5 is a
−5
→(ln 5)+
lim
( ) = ∞, so
2
= −∞. Thus,
→ (ln 5)+ . Similarly,
= ln 5 is a vertical asymptote. The graph
−5
confirms our work. 53. From the graph, it appears
3
lim →±∞
= 1 is a horizontal asymptote.
3
+ 500
3
2
+ 500
3
2
3
+ 500
= lim →±∞
+ 100 + 2000
2
3 3
+ 500
2
+ 100 + 2000 3
3 + (500
= lim →±∞
=
1 + (500
) + (100
3+ 0 = 3, 1+0+0+0
The discrepancy can be explained by the choice of the viewing window. Try
so
) 2)
+ (2000
3)
= 3 is a horizontal asymptote.
¤
117
[−100,000 100,000] by [−1 4] to get a graph that lends credibility to our calculation that
= 3 is a horizontal asymptote.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
118
¤
CHAPTER 2
LIMITS AND DERIVATIVES
54. (a)
≈ 0 and a vertical asymptote at
From the graph, it appears at first that there is only one horizontal asymptote, at
≈ 1 7. However, if we graph the function with a wider and shorter viewing rectangle, we see that in fact there seem to be ≈ 0 5 and one at √ 2 2 +1 lim ≈ 05 →∞ 3 − 5
two horizontal asymptotes: one at
≈ −0 5. So we estimate that √ 2 2 +1 lim and ≈ −0 5 →−∞ 3 − 5
(b) (1000) ≈ 0 4722 and (10,000) ≈ 0 4715, so we estimate that lim
→∞
√ 2 2 +1 ≈ 0 47. 3 −5
(−1000) ≈ −0 4706 and (−10,000) ≈ −0 4713, so we estimate that lim
→−∞
√ 2 2 +1 (c) lim = lim →∞ 3 − 5 →∞ For get
0, we have 1√ 2
2
√
+1 = −
2+1 3− 5
→−∞
[since
2
=
for
0]
√ 2 = ≈ 0 471404. 3
= = − , so when we divide the numerator by , with
2
√ 2
1 √
lim
√
2
√ 2 2 +1 ≈ −0 47. 3 −5
2
+1 = − 2+ 1
2.
0, we
Therefore,
2
√ 2 − 2+ 1 2 +1 = lim →−∞ 3− 5 3 −5
2
=−
√ 2 ≈ −0 471404. 3
55. Divide the numerator and the denominator by the highest power of
in
( ).
(a) If deg
deg
, then the numerator → 0 but the denominator doesn’t. So lim [ ( ) →∞
(b) If deg
deg
, then the numerator → ±∞ but the denominator doesn’t, so lim [ ( )
( )] = 0.
→∞
(depending on the ratio of the leading coefficients of
and
( )] = ±∞
).
56.
(i)
= 0
(ii)
0
odd)
(iii)
0
even)
(iv)
0
odd)
From these sketches we see that 1
if
=0
1
if
=0
0
if
0
(v)
0
( even)
(a) lim+ →0
=
0
if
0
∞
if
0
(b) lim
→0−
=
−∞
if
0,
odd
∞
if
0,
even
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.6
1 (c) lim
=
→∞
if
1
=0
∞
if
0
0
if
0
if
=0
if
0,
odd
∞
if
0,
even
0
if
0
−∞
(d) lim
=
→−∞
¤
LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
119
57. Let’s look for a rational function.
( )=0
⇒
degree of numerator
(2) lim ( ) = −∞
⇒
there is a factor of
(1)
lim
→±∞
→0
degree of denominator
2
in the denominator (not just , since that would produce a sign
change at = 0), and the function is negative near = 0. (3) lim ( ) = ∞ and lim+ ( ) = −∞ ⇒ vertical asymptote at →3−
= 3; there is a factor of ( − 3) in the
→3
denominator. (4)
(2) = 0
⇒
2 is an -intercept; there is at least one factor of ( − 2) in the numerator.
Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us ( )=
2− as one possibility. − 3)
2(
58. Since the function has vertical asymptotes
= 1 and
= 3, the denominator of the rational function we are looking for must
have factors ( − 1) and ( − 3). Because the horizontal asymptote is
= 1, the degree of the numerator must equal the 2
degree of the denominator, and the ratio of the leading coefficients must be 1. One possibility is ( ) = 59. (a) We must first find the function . Since
has a vertical asymptote
− 1 is a factor of the numerator. There is a removable discontinuity at
denominator and
a factor of both the numerator and denominator. Thus, be determined. Then lim
= 5. Thus ( ) =
( − 4)( + 1)
( ) = 5 lim →∞
= ( )=2
−4
→−1
3
−
4
=
2
3
−1
− 3 −4
(
= 5 lim →∞
(
2
2)
2
2
) − (1
− (3
2)
2
)
= −1, so − (−1) =
( − 1)( + 1) , where ( − 4)( + 1) (−1 − 1)
(−1 − 4)
− (4
= 5 2)
(2 − ). The -intercept is (0) = 0. The
-intercepts are 0 and 2. There are sign changes at 0 and 2 (odd exponents on and 2 −
=
=
. As
→ ∞, ( ) → −∞ because
3
→ ∞ and 2 −
→ −∞. As
1− 0 1− 0− 0
.
− 4 is a factor of the
2
, so
5
5(−1)(1) 5 = . (−4)(1) 4
→∞
60.
→−1
( − 1)
= lim
= 1,
2 5
5 − 1)( + 1) is a ratio of quadratic functions satisfying all the given conditions and ( − 4)( + 1)
2
(b) lim
now looks like this: ( ) =
( − 1)( + 1)
( ) = lim
→−1
(0) =
= 4 and -intercept
( − 1)( − 3)
= 5(1) = 5
+ 1 is
is still to = 2, and
→ −∞, of
near
( ) → −∞ because
3
→ −∞ and 2 −
= 0 flattens out (looks like
=
3
→ ∞. Note that the graph
).
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120
¤
61.
= ( )=
CHAPTER 2
LIMITS AND DERIVATIVES
−
4
6
4
=
(1 −
2
4
)=
(1 + )(1 − ).
The -intercept is
(0) = 0. The -intercepts are 0, −1, and 1 [found by solving ( ) = 0 for ]. 4
Since
0 for = −1 and
sign at
4
because 62.
= 0,
doesn’t change sign at
→ ∞ and (1 −
= ( )=
3
→ ±∞, ( ) =
= 1. As 2
= 0. The function does change 4
(1 −
2
) approaches −∞
) → −∞.
( + 2)2 ( − 1).
The -intercept is (0) = 0. The -intercepts
are 0, −2, and 1. There are sign changes at 0 and 1 (odd exponents on − 1). There is no sign change at −2. Also, ( ) → ∞ as three factors are large. And ( ) → ∞ as
→ −∞ because
( + 2) → ∞, and ( − 1) → −∞. Note that the graph of 2
=−
(looks like 63.
3
and
→ ∞ because all → −∞,
3
at
= 0 flattens out
).
= ( ) = (3 − )(1 + )2 (1 − )4 . The -intercept is (0) = 3(1)2 (1)4 = 3. The -intercepts are 3, −1, and 1. There is a sign change at 3, but not at −1 and 1. is large positive, 3 −
When
( ) = −∞. When
lim
→∞
lim
→−∞
64.
is negative and the other factors are positive, so
is large negative, 3 −
is positive, so
( ) = ∞.
= ( )=
2
(
− 1)2 ( + 2) =
2
2
( + 1)2 ( − 1)2 ( + 2).
The
-intercept is (0) = 0. The -intercepts are 0, −1, 1 and −2. There is a sign change at −2, but not at 0, −1, and 1. When positive, so lim
→∞
lim
→−∞
( ) = ∞. When
is large positive, all the factors are
is large negative, only
+ 2 is negative, so
( ) = −∞.
65. (a) Since −1 ≤ sin
≤ 1 for all
−
1
≤
→ 0. Thus, lim →∞
Theorem, (sin )
sin
sin
(b) From part (a), the horizontal asymptote is = (sin ) that is, at
=
lim
→−∞
for
0. As
= 0. = 0. The function
( ) = lim
→∞
( )=
= 0;
for every integer . Thus, the graph crosses the
66. (a) In both viewing rectangles, →∞
1
crosses the horizontal asymptote whenever sin
asymptote an infinite number of times.
lim
≤
lim
( ) = ∞ and
→−∞
( ) = −∞.
In the larger viewing rectangle, become less distinguishable.
and
→ ∞, −1
→ 0 and 1
→ 0, so by the Squeeze
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.6
( )
(b) lim →∞
5
67. lim →∞
√
→∞
( )
and
3
= lim
−5
5
3
+2
→∞
5
3
1−
= lim
5
·
1
+
2
3
2
·
LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
1
=1−
⇒
(0) + 2 (0) = 1
3
4
3
5
¤
3
have the same end behavior.
√
√
1
−1
5
= lim
·
√
1
1 − (1
→∞
5
= √
= 5 and
1− 0
)
√ − 21
10
lim →∞
1 ·
2
we have lim
10 − (21
= lim →∞
1
)
=
10 − 0
2
= 5. Since
− 21
10
2
2
( )
5 √
, −1
( ) = 5 by the Squeeze Theorem.
→∞
68. (a) After minutes, 25 liters of brine with 30 g of salt per liter has been pumped into the tank, so it contains
(5000 + 25 ) liters of water and 25 · 30 = 750 grams of salt. Therefore, the salt concentration at time will be 750
()=
5000 + 25
(b) lim
g
200 +
L
30
( ) = lim →∞
→∞
30
=
.
30
= lim →∞
200 +
200
= +
30
= 30. So the salt concentration approaches that of the brine
0+ 1
being pumped into the tank. 69. (a) lim
∗
( ) = lim
→∞
−
1−
∗
=
∗
(1 − 0) =
∗
→∞ −9 8
(b) We graph ( ) = 1 −
and ( ) = 0 99
∗
, or in this case,
( ) = 0 99. Using an intersect feature or zooming in on the point of intersection, we find that ≈ 0 47 s. 70. (a)
=
−
If (b)
−
10
1, 10
then 01
−10 ln
1 10
3
2
2
+
71. Let ( ) =
2 lim
→∞
and
( )=
3 2
= 0 1 intersect at −
⇒
−
10
ln 0 1
= −10 ln 10 +1
≈ 23 03.
0 1.
−1
⇒
= 10 ln 10 ≈ 23 03
and ( ) = ( ) − 1 5 . Note that
+1
and lim
→∞
-value at which ( ) so we choose
10
1
( ) = 0. We are interested in finding the 0 05. From the graph, we find that
= 15 (or any larger number).
≈ 14 804,
121
72. We want to find a value of
such that
⇒
1− 3 − (−3) √ 2 +1 1− 3
1− 3 −3 −
√
2
+1
, or equivalently,
−3 + . When
= 0 1, we graph
= ( )= √
2
,
= −3 1, and
= −2 9. From the graph,
+1
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
122
¤
CHAPTER 2
LIMITS AND DERIVATIVES
we find that ( ) = −2 9 at about that ( ) = −2 95 at about
73. We want a value o
= 11 283, so we choose
= 21 379, so we choose
1− 3 ⇒ √
such that
2
= 12 (or any larger number). Similarly for
= 0 05, we find
= 22 (or any larger number).
−3
1− 3 √
, or equivalently, 3 −
2
+1
3 + . When
= 0 1,
+1
choose
1− 3 = ( )= √ 2 , = 3 1, and = 2 9. From the graph, we find that ( ) = 3 1 at about = −8 092, so we +1 = −9 (or any lesser number). Similarly for = 0 05, we find that ( ) = 3 05 at about = −18 338, so we
choose
= −19 (or any lesser number).
we graph
74. We want to find a value of
We graph
= ( )=
√
ln
that ( ) = 100 at about
√
⇒
such that
ln
100.
= 100. From the graph, we find
and
= 1382 773, so we choose
= 1383 (or
any larger number). 75. (a) 1
2
0 0001
(b) If
⇔
2
1 0 0001 = 10 000
1 √
⇒
Then
⇔
2
0 is given, then 1
2
1
⇒
√
0 0001
(b) If
0 is given, then 1
Then 77. For
Take
78. Given
√
⇔
−0 =
0, 1
⇔
1
⇒ − 0 = −1
1
√
0)
. Le 1
, so lim →∞
⇔ 1
√
(
=1
√
.
= 0.
2
108 ⇔
1
2
. Let
=1
2
.
1 , so lim √ = 0. →∞
. If
0 is given, then −1
= −1 . Then
⇒
0, we need √ 3 = ⇒
0 such that 3
1
1 1 √ −0 = √
⇒
2
⇔
2
1 0 0001 = 104
√
100
1
2
76. (a) 1
⇔
−1
, so lim
→∞
⇒
⇒ 3
= ∞.
(1
3
⇔
−1 .
) − 0 = −1
. Now
3
, so lim (1 →−∞
⇔
√ 3
) = 0.
, so tak
=
√ 3
. Then
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.7
79. Given
0, we need
= max(1 ln
⇔
. Now
0.) Then
= max(1 ln
)
ln
⇒
→∞
80. Definition
number
Let
be a function defined on some interval (−∞ ). Then lim
→−∞
there is a corresponding negative number
prove that lim
→−∞
3
1+
→−∞
whenever
( ) = . Then for every
. If = 1
corresponding lim
, so take )≥
max(
,
= lim
→∞
whenever
= lim
→0+
=
√ 3
0
1
⇔
1
0
1
sin
1
= lim
sin
whenever 0
1
sin
0 there is a corresponding negative number ⇔
, then
0 (namely −1
1
1
) such that (1 ) −
0
⇔
1
whenever −

( ). 1
[let
= ]
[part (a) with
=1 ]
→∞
= lim
→∞
= 0
sin
such that ( ) − 0 there is a
. This proves that
→0+
= lim
− 1 and find that
. Thus, for every
→−∞
→0−
(b) lim
− 1. Thus, we tak
⇒
such that
= −∞.
3
. Then for every
. If = 1
0 there is a corresponding (1 ) =
√ 3
) such that (1 ) −
( )=
→−∞
lim
. Now we use the definition to
( ).
Now suppose that lim
whenever
, we need a negative numbe
⇔
1+
( ) = −∞ means that for every negative
0 there is a corresponding positive number
⇔
, then
0 (namely 1
(1 ) =
( )−
−1
3
. This proves that lim
→∞
→0+
⇔
3
3
81. (a) Suppose that lim
such that ( )
= −∞. Given a negative number
3
. Now 1 + ⇒ 1+
2.7
123
= ∞.
so lim
1+
⇒
0 such that
). (This ensures that
¤
DERIVATIVES AND RATES OF CHANGE
[let
= ]
[by Exercise 65]
Derivatives and Rates of Change
1. (a) This is just the slope of the line through two points:
(b) This is the limit of the slope of the secant line
a
=
( ) − (3) . −3
∆ = ∆
approaches
2. The curve looks more like a line as the viewing rectangle gets smaller.
:
= lim
→3
( ) − (3) . −3
such that 0. Thus, for every 0. This proves that
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124
¤
CHAPTER 2
LIMITS AND DERIVATIVES
3. (a) (i) Using Definition 1 with ( ) = 4 −
( )− ( )
= lim
and
(1 3),
(4 −
= lim
−
→
2
2
)−3
−1
→1
−(
= lim
2
− 4 + 3) −1
→1
−( − 1)( − 3)
= lim
−1
→1
= lim (3 − ) = 3 − 1 = 2 →1
(ii) Using Equation 2 with ( ) = 4 −
2
( + )− ( )
= lim
and
(1 3), (1 + ) − (1)
= lim
→0
→0
→0 2
2
= lim
4+ 4 − 1− 2 −
−3
−
= lim
→0
+2
0
(− + 2)
= lim
→0
(b) An equation of the tangent line is − ( ) = or
4(1 + ) − (1 + ) 2 − 3
= lim
= lim (− + 2) = 2
→0
( )( − ) ⇒
→0
− (1) =
0
(1)( − 1)
⇒
− 3 = 2( − 1),
= 2 + 1.
(c)
The graph of
=4 −
= 2 + 1 is tangent to the graph of
2
at the
point (1 3). Now zoom in toward the point (1 3) until the parabola and the tangent line are indistiguishable.
4. (a) (i) Using Definition 1 with ( ) =
−
3
and
( )− 0
= lim
= lim
3
and
( + )− ( )
→0
= lim
(1 + ) − (1)
→0
− (1 + 3 + 3
2
(c)
(1 + )(1 − ) −1
→1
+
(1 + ) − (1 + )3 − 0
= lim 3
)
= lim
−
2
−3
2
−2
→0
= lim
(−
− 3 − 2)
→0
− 3 − 2) = −2
(b) An equation of the tangent line is or
= lim
→0
3
→0
= lim (−
)
−1
→1
→0
1+
2
(1 0),
2
= lim
= lim
] = −1(2) = −2
→1
−
(1 −
3
−1
→1
= lim [− (1 + (ii) Using Equation 2 with ( ) =
−
= lim
−1
→1
(1 0),
− ( )=
0
( )( − ) ⇒
− (1) =
0
(1)( − 1)
⇒
− 0 = −2( − 1),
= −2 + 2. The graph of
= −2 + 2 is tangent to the graph of
=
−
3
at the
point (1 0). Now zoom in toward the point (1 0) until the cubic and the tangent line are indistinguishable.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.7
5. Using (1) with ( ) = 4 − 3
2
(2 −4) [we could also use (2)],
and
( )− ( )
= lim
Tangent line:
3
⇔
+ 4 = −8 + 16
− 3 + 1 and
8 + 12 + 6
→0 2
− 6− 3 − 2
+
2
→0
√ − 1
√ = lim
1 2(
8. Using (1) with ( ) =
− 3 = 9 − 18 (
= lim
= lim
⇔
1 2
=
2 +1 and +2
⇔
1
= lim →1
( − 1)
1
( − 1)(
→1
+ 1)
−1 =
−
1
−2
2
3
( + )− ( )
+2 −1
→1
−1
= lim →1
( − 1)( + 2)
⇔
1
=
3
1
+
3
2 3
= lim
3 + 4( + )2 − 2( + )3 − (3 + 4
−2
2
3
3 + 4(
2
+2
+
3
) − 2(
2
+3
2
+3
3
2
)− 3− 4 −
+
3
→0
3
+2 2
2
3+ 4
2
+8
+4
2
−2
− 63
3
−6
2
−2
− 3− 4
+2
3 2
→0
→0
)
→0
8
.
2
+1
,
2
= lim
1
=
2 + 1 − ( + 2)
1 = lim
→0
= lim
1
= lim √
3
⇔
= ( )= 3 +4
= lim
−1 √
1
=
1+ 2
−
3
= lim
→1
−1
→1
=
+2
)
1 2
+2
3
9. (a) Using (2) with
= lim
= 9 − 15
+ 1)
+
= lim
−
→
1
2
+6 +
(1 1),
( )− ( )
= lim
−1 =
+
→0
√ − 1)( + 1) √ = lim
( − 1)(
→1
− 1)
√
2 +1
Tangent line:
3
9 +6
)=9
⇔
−1
→1
−1 =
(2 + )3 − 3(2 + ) + 1 − 3
→0
= lim (9 + 6 +
Tangent line:
= lim
3
→0
7. Using (1),
= −8 + 12.
→0
− 3 = 9( − 2)
+4 +4 −2
→2
⇔
(2 + ) − (2)
= lim
2
Tangent line:
2
(2 3),
→0
= lim
−3
= lim
−2
→2
( + )− ( )
= lim
− (−4)
2
(−3 − 2)( − 2) = lim (−3 − 2) = −3(2) − 2 = −8 →2 −2
− (−4) = −8( − 2)
6. Using (2) with ( ) =
4 −3
= lim
−
→
= lim →2
DERIVATIVES AND RATES OF CHANGE
+4
2
−6
2
−6
2
−2
= lim →0
(8 + 4
6
2
−6
−2
)
¤
125
= lim (8 + 4 − 6 →0
(b) At (1 5):
−6
−2
2
)=8 −6
= 8(1) − 6(1)2 = 2, so an equation of the tangent line
is − 5 = 2( − 1) At (2 3):
2
⇔
2
(c)
= 2 + 3.
= 8(2) − 6(2)2 = −8, so an equation of the tangent
line is − 3 = −8( − 2)
⇔
= −8 + 19.
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126
¤
CHAPTER 2
LIMITS AND DERIVATIVES
10. (a) Using (1),
1 √
√
1 − √
−
√
√
= lim
(
= lim √
= lim −
→
→
(
−
→
−1
= lim √
√
+
)
1
=−
√ 2 (2 )
−
2
√
√ √ )( + )
or −
3 2
1
−3 2
is − 1 = − ( − 1)
⇔
1
2
=−
1
[
√ ( − )(
√ +
)
0]
2
= − 21 , so an equation of the tangent line
(b) At (1 1):
−
√ √ = lim √ → ( − )( + )
→
−1
= √
√
√
(c)
3
+ .
2
2
1 = − 16 , so an equation of the tangent line
At 4
1 2
:
is −
1 2
1 = − 16 ( − 4)
⇔
1 = − 16
11. (a) The particle is moving to the right when
+ 34 .
is increasing; that is, on the intervals (0 1) and (4 6). The particle is moving to
the left when is decreasing; that is, on the interval (2 3). The particle is standing still when is constant; that is, on the intervals (1 2) and (3 4). (b) The velocity of the particle is equal to the slope of the tangent line of the graph. Note that there is no slope at the corner points on the graph. On the interval (0 1) the slope is
3− 0 = 3. On the interval (2 3), the slope is 1− 0
1− 3 3− 1 = −2. On the interval (4 6), the slope is = 1. 3− 2 6− 4
12. (a) Runner A runs the entire 100-meter race at the same velocity since the slope of the position function is constant.
Runner B starts the race at a slower velocity than runner A, but finishes the race at a faster velocity. (b) The distance between the runners is the greatest at the time when the largest vertical line segment fits between the two graphs—this appears to be somewhere between 9 and 10 seconds. (c) The runners had the same velocity when the slopes of their respective position functions are equal—this also appears to be at about 9 5 s. Note that the answers for parts (b) and (c) must be the same for these graphs because as soon as the velocity for runner B overtakes the velocity for runner A, the distance between the runners starts to decrease. 13. Let ( ) = 40 − 16 2 .
(2) = lim →2
= lim →2
( ) − (2) −2
= lim →2
40 − 16
2
−2
− 16
= lim →2
−16
2
+ 40 − 16 −2
−8( − 2)(2 − 1) = −8 lim(2 − 1) = −8(3) = −24 →2 −2
Thus, the instantaneous velocity when = 2 is −24 ft s.
= lim →2
−8 2
2
− 5 +2 −2
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.7
14. (a) Let
DERIVATIVES AND RATES OF CHANGE
¤
127
( ) = 10 − 1 86 2 . (1 + ) −
(1) = lim
(1)
10(1 + ) − 1 86(1 + ) 2 − (10 − 1 86)
= lim
→0
→0
10 + 10 − 1 86(1 + 2 +
= lim
) − 10 + 1 86
2
→0
10 + 10 − 1 86 − 3 72 − 1 86
= lim
2
− 10 + 1 86
→0
6 28 − 1 86
= lim
2
= lim (6 28 − 1 86 ) = 6 28 →0
→0
The velocity of the rock after one second is 6 28 m s. ( + )−
(b) ( ) = lim
( )
= lim
→0
= lim
10( + ) − 1 86( + )2 − (10 − 1 86
2
)
→0
10 + 10 − 1 86(
2
+2
2
+
) − 10 + 1 86
2
→0 2
= lim
10 + 10 − 1 86
2
− 3 72
− 1 86
2
2
− 10 + 1 86
10 − 3 72
= lim
→0
− 1 86
→0
(10 − 3 72 − 1 86 )
= lim
→0
= lim (10 − 3 72 − 1 86 ) = 10 − 3 72 →0
is (10 − 3 72 ) m s
The velocity of the rock when = (c) The rock will hit the surface whe
⇔
= 0
10 − 1 86
2
= 0
⇔
(10 − 1 86 ) = 0
⇔
= 0 or 1 86 = 10.
The rock hits the surface when = 10 1 86 ≈ 5 4 s.
1 15.
( + )− ( )
( ) = lim
= lim
→0
2
−(2
→0
−2 13
2(
+
2
)
+ )2
= lim
= −2 m s, (2) =
( + )2
2
= lim
2(
−2
= lim
=−
23
1
2(
+ )2
m s, and (3) =
=
2
=
−2 2
·
= 2
+2
2(
−2
m s
3
2 = − m s. 27
( + ) − 6( + ) + 23 − 2
+
+
− 6 + 23 1 2 2
( + )− 1 2 2
2
is
1
=
−2 33
4
16. (a) The average velocity between times and +
( + )− ( )
−(2 + )
→0
+ )2
−(
→0
→0
− (2 + )
→0
= 10 − 20 = −10 m s.
10 1 86
− ( + )2 2
= lim
→0
= lim
So (1) =
2
1
−
( + )2
= 10 − 3 72
10 1 86
(d) The velocity of the rock when it hits the surface is
1 2
2
− 6 − 6 + 23 −
1 2 2
+ 6 − 23
+
+ )2
2
)
+ =
1 2
2
−6
+
1 2
=
−6 =
(i) [4 8]: = 4,
= 8 − 4 = 4, so the average velocity is 4 +
1 2 (4)
− 6 = 0 ft s.
(ii) [6 8]: = 6,
= 8 − 6 = 2, so the average velocity is 6 +
1 2 (2)
− 6 = 1 ft s.
+
(iii) [8 10]: = 8,
= 10 − 8 = 2, so the average velocity is 8 +
1 2 (2)
− 6 = 3 ft s.
(iv) [8 12]: = 8,
= 12 − 8 = 4, so the average velocity is 8 +
1 2 (4)
− 6 = 4 ft s.
1 2
− 6 ft s
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128
¤
CHAPTER 2
LIMITS AND DERIVATIVES
( + )− ( )
(b) ( ) = lim
= lim
= − 6,
17.
0
so (8) = 2 ft s.
(0) is the only negative value. The slope at
at
(c)
2
→0
→0
−6
1
+
= −2. Thus,
0
(0)
0
0
0
(4)
= 4 is smaller than the slope at 0
(2)
= 2 and both are smaller than the slope
(−2).
(60) − (20) 700 − 300 400 = = = 10 40 40 60 − 20
18. (a) On [20 60]:
(b) Pick any interval that has the same -value at its endpoints. [0 57] is such an interval since (0) = 600 and (57) = 600. (60) − (40) 700 − 200 500 = = = 25 20 20 60 − 40
(c) On [40 60]:
(70) − (40) 900 − 200 700 = = = 23 13 70 − 40 30 30
On [40 70]:
23 13 , the average rate of change on [40 60] is larger.
Since 25 (d)
(40) − (10) 200 − 400 −200 = = = −6 23 40 − 10 30 30 This value represents the slope of the line segment from (10
19. (a) The tangent line at
(50) ≈
60 − 43
(b) The tangent line at value, that is,
0
440 =
17
≈ 26.
= 10 is steeper than the tangent line at
(10)
0
0
(60)
= 30, so it is larger in magnitude, but less in numerical
(30).
(c) The slope of the tangent line at So yes,
= 60,
0
(60), is greater than the slope of the line through (40
0
(5) = 4, the slope of the tangent line at
Using the point-slope form of a line gives us − (−3) = 4( − 5), or = 4 − 5: when
= 2,
tangency, these values are shared with the curve 22. Since (4 3) is on
(40)) and (80
(80)).
(80) − (40) . 80 − 40
20. Since (5) = −3, the point (5 −3) is on the graph of . Since
21. For the tangent line
(40)).
= 50 appears to pass through the points (43 200) and (60 640), so
640 − 200 0
(10)) to (40
= 5 is 4.
= 4 − 23.
= 4(2) − 5 = 3 and its slope is 4 (the coefficient of ). At the point of = ( ); that is, (2) = 3 and
0
(2) = 4.
= ( ), (4) = 3. The slope of the tangent line between (0 2) and (4 3) is 14, so
0
(4) = 14.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.7
DERIVATIVES AND RATES OF CHANGE
23. We begin by drawing a curve through the origin with a 0
slope of 3 to satisfy (0) = 0 and 0
(0) = 3. Since
(1) = 0, we will round off our figure so that there is
a horizontal tangent directly over
= 1. Last, we
make sure that the curve has a slope of −1 as we pass over
= 2. Two of the many possibilities are shown.
24. We begin by drawing a curve through the origin with a slope of 1 to satisfy
(0) = 0 and
0
(0) = 1. We round off our figure at
and then pass through (2 0) with slope −1 to satisfy (2) = 0 and We round the figure at
= 3 to satisfy 0
with slope 1 to satisfy (4) = 0 and
0
0
(1) = 0,
(2) = −1.
(3) = 0, and then pass through (4 0)
(4) = 1 Finally we extend the curve on
( ) = ∞ and lim
both ends to satisfy lim
0
= 1 to satisfy
→∞
( ) = −∞.
→−∞
25. We begin by drawing a curve through (0 1) with a slope of 1 to satisfy (0) = 1
and
0
= −2 to satisfy
(0) = 1. We round off our figure at
→ −5+ ,
→ ∞, so we draw a vertical asymptote at
0
(−2) = 0. As
= −5. As
→ 5− ,
→ 3, so we draw a dot at (5 3) [the dot could be open or closed].
26. We begin by drawing an odd function (symmetric with respect to the origin)
through the origin with slope −2 to satisfy 0 (0) = −2. Now draw a curve starting at = 1 and increasing without bound as → 2− since lim− ( ) = ∞. Lastly, →2
reflect the last curve through the origin (rotate 180◦ ) since
27. Using (4) with ( ) = 3 0
2
−
(1) = lim
3
and
= 1,
(1 + ) − (1)
= lim
[3(1 + )2 − (1 + )3 ] − 2
→0
→0
= lim
is an odd function.
(3 + 6 + 3
2
) − (1 + 3 + 3
2
+
3
)−2
→0
= lim (3 − →0
Tangent line:
− 2 = 3( − 1)
= lim →0
2
3 −
3
= lim
(3 −
2
)
→0
) = 3 −0 = 3
⇔
−2 = 3 −3
⇔
=3 −1
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¤
129
130
¤
CHAPTER 2
LIMITS AND DERIVATIVES 4
28. Using (5) with ( ) = 0
− 2 and
(1) = lim
2
(
−1
+ 1)( + 1)( − 1)
⇔ 2
(1 +
(2 + ) −
(2) = lim
(
(2)
2
2
− 1)
−1
→1 2
+ 1)( + 1)] = 2(2) = 4
+1 =4 − 4
⇔
=4 −5
) and the point (2 2), we have
1 + (2 + )
= lim
→0
+ 1)(
→1
5(2 + ) 0
= lim
−1
→1
= lim [(
−1
( )=5
−1
= lim 4
− (−1) = 4( − 1)
29. (a) Using (4) with
− 2) − (−1)
→1
→1
Tangent line:
4
(
= lim
−1
→1
= lim
= 1,
( ) − (1)
(b)
−2
2
→0
5 + 10 2
= lim
+4 +5
−
5 + 10 − 2(
2
2
= lim
→0
→0
2
−2
= lim →0
2
(
−3
→0
2
(
( )=4 0
2
−
3
2
=
+4 +5
−3 5
= − 35 +
16 5.
, we have ( )
= lim
→0
[4( + ) 2 − ( + ) 3 ] − (4
2
−
3
)
→0 2
= lim
→0
− 2 = − 35 ( − 2) or
( + )−
( ) = lim
−2 − 3
= lim
+ 4 + 5)
So an equation of the tangent line at (2 2) is 30. (a) Using (4) with
+ 4 + 5)
+4 +5
(−2 − 3)
= lim
+ 4 + 5)
2
2
4
+8
8
+4
3
−(
+4
2
2
+3
3
+3
+
2
)−4
3
+
→0
= lim
2
−3
−3
2
2
−
= lim
→0 2
→0
0
0
−3
−
2
)=8 −3
(2) = 16 − 12 = 4, and an equation of the
tangent line is − 8 = 4( − 2), or
2
−3
−
→0
= lim (8 + 4 − 3 At the point (2 8),
(8 + 4 − 3
3
2
(b)
= 4 . At the point (3 9),
(3) = 24 − 27 = −3, and an equation of the tangent line is
− 9 = −3( − 3), or
31. Use (4) with ( ) = 3 0
2
= −3 + 18
− 4 + 1.
( ) = lim
( + )− ( )
→0
= lim
[3( + )2 − 4( + ) + 1] − (3
2
− 4 + 1)]
→0 2
2
2
2
2
)
= lim
3
+6
+3
− 4 − 4 +1 −3
+4 − 1
→0
= lim
→0
= lim
6
+3
−4
→0
(6 + 3 − 4)
= lim (6 + 3 − 4) = 6 − 4 →0
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.7
32. Use (4) with ( ) = 2 0
3
DERIVATIVES AND RATES OF CHANGE
+ . ( + )− ( )
( ) = lim →0
2
2
2
+ )
+6
3
3
+6
+2
−
−2
+ +
2
2
6
= lim
→0
+6
3
+2
→0 2
= lim
3
→0 3
= lim
[2( + )3 + ( + )] − (2
= lim
2
+6
(6
+2
+ 1)
2
= lim (6
→0
→0
+6
+2
2
2
+ 1) = 6
+1
33. Use (4) with ( ) = (2 + 1) ( + 3). 0
( + )− ( )
( ) = lim
= lim
→0
= lim
→0
= lim
→0
(2 + 2 + 1)( + 3) − (2 + 1)( + ( + + 3)( + 3) 2
(2
+ 6 + + 3) − (2
+6 +2
→0
( + 5
= lim →0
34. Use (4) with ( ) =
−2
0
2
=1
( +
+2
+6 + +
+ 3)
( +
5
=
( + 3)2
+ 3)( + 3)
.
( ) = lim
( + )− ( )
−(
→0
= lim →0
1 1 − 2 ( + )2
= lim
2
2
+2
2(
2
)
→0
= lim →0
+ )2
−2 − 2(
+
+ )2
−2
−
2
= lim
2(
+ )2
1 − 2( + ) −
√ 1− 2
−2
=
− ( + )2 + )2
2(
= lim
→0
2
= lim
0
2
5
→0
+ 3)( + 3)
+ 3)
+ 3)( + 3)
= lim
→0
35. Use (4) with ( ) =
2( + ) + 1 2 +1 − ( + )+3 +3
=
→0
(−2 − ) 2(
+ )2
−2
2( 2)
3
√ 1− 2 .
( ) = lim
( + )− ( )
= lim
→0
→0
1 − 2( + ) −
√ 1− 2
= lim
√ 1− 2
1 − 2( + ) +
√ 1− 2
·
→0
√
2
1 − 2( + ) = lim
1 − 2( + ) +
−
2
1− 2 = lim
(1 − 2 − 2 ) − (1 − 2 )
+
¤
131
→0
= lim →0
1 − 2( + ) +
√ 1−2
−2 1 − 2( + ) +
→0
= lim √ 1−2
1 − 2( + ) +
√ 1− 2
−2 1 − 2( + ) +
√ 1− 2
→0
= √
−2 −2 −1 = = √ √ √ 1− 2 1− 2 + 1− 2 2 1− 2
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132
¤
CHAPTER 2
LIMITS AND DERIVATIVES
4 . 1−
36. Use (4) with ( ) = √
0
( + )− ( )
( ) = lim
4 4 − √ 1− ( + ) 1−
= lim
→0
→0
√ √ 1− − 1− − √ √ 1− − 1−
√ 1−
= 4 lim
= 4 lim →0
→0
√ 1− = 4 lim
√ 1−
→0
= 4 lim →0
√ 1−
−
−
=
√ 1−
√ 1−
·√ 1−
(1 − ) − (1 − √ 1−
√ √ 1− ( 1−
−
1 √ √ 1− ( 1−
= 4 lim √ →0
−
1−
−
−2+
38. By (4), lim
0
=
→0 6
39. By Equation 5, lim
− 64
1 →1 4
cos( + ) + 1
0
=
0
=
sin −
0
(4) = lim
=
2 (1 − )3
−
−
1 √ 1− ( 1−
√
2
and
(2), where ( ) =
6
= 9.
and
=
2. −
and
= 2.
6
(4), where ( ) =
1
and
=
−
1
.
1 2
0
( ), where ( ) = cos
=
0
=
0
and
= .
(0), where ( ) = cos( + ) and
6
, where ( ) = sin
and
=
= 0.
6
.
6
(4 + ) − (4)
= lim
80(4 + ) − 6(4 + )2 − 80(4) − 6(4)2
− )2 √
√ 1− ( 1−
√ √ 1− ( 1−
4
cos( + ) + 1
→
2
√ 1−
→0
= 4·√ 1− − )
1−
√
1−
→0
− )
√ +
√
= 4 lim
( 2), where ( ) = −
=
→0
(4) =
− ) √ + 1−
−
1 − 4
→0
42. By Equation 5, lim
√ √ ( 1 − )2 − ( 1 − = 4 lim
1−
−4
40. By Equation 5, lim
Or: By (4), lim
√ 1−
−2
→2
41. By (4), lim
−
(9), where ( ) =
−2
−
√ 1−
−
√ 0
=
→0
−
+
−3
37. By (4), lim
√ 1−
√
2 4 √ = (1 − )1 (1 − )1 (1 − )(2 1 − ) √ 9+
43.
+
√ 1−
−
+
+
√ +
1− )
1−
√ 1−
− )
− )
→0
= lim
→0
(320 + 8
− 96 − 48 − 6
) − (320 − 96)
2
→0
= lim
→0
= lim
32 − 6
2
→0
(32 − 6 )
= lim (32 − 6 ) = 32 m/s →0
The speed when = 4 is 32 = 32 m s.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.7
45
10 + 44.
(4) =
0
(4 + ) − (4)
(4) = lim →0
= lim
4+
= lim
DERIVATIVES AND RATES OF CHANGE
45
10 + +1
4+1
−
→0
45 − 9(5 + )
→0
= lim →0
(5 + )
The speed when = 4 is
− 95
=
45 = lim 5 +
¤
133
9 −
→0
−9
−9
= lim
(5 + )
→0
9
=−
5+
m/s.
5
m s. 9 5
45. The sketch shows the graph for a room temperature of 72◦ and a refrigerator
temperature of 38◦ . The initial rate of change is greater in magnitude than the rate of change after an hour.
46. The slope of the tangent (that is, the rate of change of temperature with respect
to time) at = 1 h seems to be about
47. (a) (i) [1 0 2 0]:
(2) − 2− 1
(1)
=
75 − 168 ◦ ≈ −0 7 F min. 132 − 0
0 18 − 0 33
= −0 15
1
mg/mL h
(ii) [1 5 2 0]:
(2) − (1 5) 0 18 − 0 24 −0 06 mg/mL = = = −0 12 05 05 h 2− 15
(iii) [2 0 2 5]:
(2 5) − (2) 0 12 − 0 18 −0 06 mg/mL = = = −0 12 05 05 h 25− 2
(iv) [2 0 3 0]:
(3) − (2) 0 07 − 0 18 mg/mL = = −0 11 3− 2 1 h
(b) We estimate the instantaneous rate of change at = 2 by averaging the average rates of change for [1 5 2 0] and [2 0 2 5]: mg/mL −0 12 + (−0 12) = −0 12 . After 2 hours, the BAC is decreasing at a rate of 0 12 (mg mL) h. h 2
48. (a) (i) [2006 2008]:
(ii) [2008 2010]:
(2008) − (2006) 16,680 − 12,440 4240 = = = 2120 locations year 2008 − 2006 2 2 (2010) − (2008) 16,858 − 16,680 178 = = = 89 locations year. 2010 − 2008 2 2
The rate of growth decreased over the period from 2006 to 2010. (b) [2010 2012]:
(2012) − (2010) 18,066 − 16,858 1208 = = = 604 locations year. 2012 − 2010 2 2
Using that value and the value from part (a)(ii), we have
89 + 604 693 = = 346 5 locations year. 2 2
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134
¤
CHAPTER 2
LIMITS AND DERIVATIVES
(c) The tangent segment has endpoints (2008 16,250) and (2012 17,500). An estimate of the instantaneous rate of growth in 2010 is 17,500 − 16,250 1250 = = 312 5 locations/year. 2012 − 2008 4
49. (a) [1990 2005]:
84,077 − 66,533 17,544 = = 1169 6 thousands of barrels per day per year. This means that oil 2005 − 1990 15
consumption rose by an average of 1169 6 thousands of barrels per day each year from 1990 to 2005. 76,784 − 70,099 6685 = = 1337 2000 − 1995 5
(b) [1995 2000]:
84,077 − 76,784 7293 = = 1458 6 2005 − 2000 5
[2000 2005]:
An estimate of the instantaneous rate of change in 2000 is
1 2
(1337 + 1458 6) = 1397 8 thousands of barrels
per day per year. (11) −
50. (a) (i) [4 11]:
(4)
=
9 4 − 53
11 − 4 (11) −
(ii) [8 11]:
=
−43 6
7 (8)
=
9 4 − 18
11 − 8
≈−
7
=
−8 6
3
3
≈−
6 23
RNA copies mL day
2 87
RNA copies mL day
(15) − (11) 52− 94 −4 2 RNA copies mL = = = −1 05 day 15 − 11 4 4
(iii) [11 15]:
(22) −
(iv) [11 22]:
(11)
22 − 11
0
(b) An estimate of
=
36− 94
=
11
−5 8 11
≈−
0 53
RNA copies mL day
(11) is the average of the answers from part (a)(ii) and (iii).
(11) ≈
0
(11) measures the instantaneous rate of change of patient 303’s viral load 11 days after ABT-538 treatment began.
1 2
[−2 87 + (−1 05)] = −1 96
RNA copies mL . day
0
51. (a) (i)
∆ ∆
=
(105) − (100) 6601 25 − 6500 = = $20 25 unit. 105 − 100 5
(ii)
∆ ∆
=
(101) − (100) 6520 05 − 6500 = = $20 05 unit. 101 − 100 1
(b)
(100 + ) −
(100)
=
5000 + 10(100 + ) + 0 05(100 + ) 2 − 6500
= 20 + 0 05 , So the instantaneous rate of change is lim
→0
=
20 + 0 05
2
=0 (100 + ) −
(100)
= lim (20 + 0 05 ) = $20 unit. →0
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.7 2
+ 52. ∆
=
( + )−
1−
=
100,000 3600
Dividing ∆
30
+
( 120 + 2 + ) = −
by
− 100,000 1 −
60
−
250 9
1−
30
+
3600
2
2 = 100,000 −
30
+
3600
+
3600
( 120 + 2 + ) −
→ 0, we see that the instantaneous rate of change is
and then letting
135
60
2
3600
¤
2
( + )2
+ = 100,000
1−
( ) = 100,000
DERIVATIVES AND RATES OF CHANGE
Flow rate (gal min)
Water remaining
0
−3333 3
10
−2777 7
69 444 4
20
−2222 2
44 444 4
30
−1666 6
25 000
40
−1111 1
11 111 1
50
− 555 5
2 777 7
( − 60) gal min.
( ) (gal)
100 000
0
60
500 9
0
The magnitude of the flow rate is greatest at the beginning and gradually decreases to 0. 53. (a)
0
( ) is the rate of change of the production cost with respect to the number of ounces of gold produced. Its units are
dollars per ounce. (b) After 800 ounces of gold have been produced, the rate at which the production cost is increasing is $17 ounce. So the cost of producing the 800th (or 801st) ounce is about $17. (c) In the short term, the values of eventually 54. (a)
0
0
0
( ) will decrease because more efficient use is made of start-up costs as
increases. But
( ) might increase due to large-scale operations.
(5) is the rate of growth of the bacteria population when = 5 hours. Its units are bacteria per hour.
(b) With unlimited space and nutrients,
0
should increase as increases; so
0
(5)
0
(10). If the supply of nutrients is
limited, the growth rate slows down at some point in time, and the opposite may be true. 0
55. (a)
(58) is the rate at which the daily heating cost changes with respect to temperature when the outside temperature is ◦
58 F. The units are dollars
◦
F.
(b) If the outside temperature increases, the building should require less heating, so we would expec 56. (a)
0
0
(58) to be negative.
(8) is the rate of change of the quantity of coffee sold with respect to the price per pound when the price is $8 per pound.
The units for (b)
0
0
(8) are pounds (dollars pound).
(8) is negative since the quantity of coffee sold will decrease as the price charged for it increases. People are generally
less willing to buy a product when its price increases. 57. (a)
0
( ) is the rate at which the oxygen solubility changes with respect to the water temperature. Its units are (mg L)
(b) For 0
◦
C.
= 16◦ C, it appears that the tangent line to the curve goes through the points (0 14) and (32 6). So
(16) ≈
6 − 14 8 =− = −0 25 (mg L) ◦ C. This means that as the temperature increases past 16◦ C, the oxygen 32 32 − 0
solubility is decreasing at a rate of 0 25 (mg L)
◦
C.
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136
¤
CHAPTER 2 0
58. (a)
LIMITS AND DERIVATIVES
( ) is the rate of change of the maximum sustainable speed of Coho salmon with respect to the temperature. Its units
are (cm s) ◦ C. = 15◦ C, it appears the tangent line to the curve goes through the points (10 25) and (20 32). So
(b) For 0
(15) ≈
32 − 25 = 0 7 (cm s) ◦ C. This tells us that at 20 − 10
= 15◦ C, the maximum sustainable speed of Coho salmon is
changing at a rate of 0.7 (cm s) ◦ C. In a similar fashion for 0
obtain
(25) ≈
= 25◦ C, we can use the points (20 35) and (25 25) to
25 − 35 = −2 (cm s) ◦ C. As it gets warmer than 20◦ C, the maximum sustainable speed decreases 25 − 20
rapidly. 59. Since ( ) = 0
(0) = lim
sin(1
) when
(0 + ) − (0)
→0
= 0 and (0) = 0, we have = lim
sin(1
)− 0
→0
= lim sin(1 →0
). This limit does not exist since sin(1
) takes the
values −1 and 1 on any interval containing 0. (Compare with Example 2.2.4.) 60. Since ( ) = 0
(0) = lim
2
sin(1
) when
(0 + ) − (0)
→0
= 0 and (0) = 0, we have 2
= lim →0
lim
→0
sin
1
= lim
sin(1
). Since −1 ≤ sin
1
≤ 1, we have
→0
1 − ≤ sin
)− 0
sin(1
1 ≤
⇒
− ≤
sin
≤ . Because lim (− ) = 0 and lim = 0, we know that
= 0 by the Squeeze Theorem. Thus,
61. (a) The slope at the origin appears to be 1.
(b) The slope at the origin still appears to be 1.
(c) Yes, the slope at the origin now appears to be 0.
→0
0
(0) = 0.
→0
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SECTION 2.8
2.8
¤
THE DERIVATIVE AS A FUNCTION
137
The Derivative as a Function
1. It appears that 0
is,
(− ) =
is an odd function, so 0
0
will be an even function—that
( ).
(a)
0
(−3) ≈ −0 2
(b)
0
(−2) ≈ 0
(e)
0
(1) ≈ 1
(c)
0
(−1) ≈ 1
(f )
0
(2) ≈ 0
(d)
0
(0) ≈ 2
(g)
0
(3) ≈ −0 2
2. Your answers may vary depending on your estimates.
(a) Note: By estimating the slopes of tangent lines on the graph of , it appears that (b)
0
(1) ≈ 0
(c)
0
(2) ≈ −1 5
(f )
0
(5) ≈ −0 3
0
(0) ≈ 6.
(d)
0
(3) ≈ −1 3
(g)
0
(6) ≈ 0
(e)
0
(4) ≈ −0 8
(h)
0
(7) ≈ 0 2
3. (a)0 = II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0, then
negative again. The actual function values in graph II follow the same pattern. 0
(b) = IV, since from left to right, the slopes of the tangents to graph (b) start out at a fixed positive quantity, then suddenly become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents. (c)0 = I, since the slopes of the tangents to graph (c) are negative for
0 and positive for
0, as are the function values of
graph I. 0
(d) = III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then positive, then 0, then negative again, and the function values in graph III follow the same pattern. Hints for Exercises 4 –11: First plot -intercepts on the graph of of — there will be a discontinuity on the graph of
0
0
for any horizontal tangents on the graph of . Look for any corners on the graph
. On any interval where
positive (or negative). If the graph of the function is linear, the graph of
4.
0
has a tangent with positive (or negative) slope, the graph of
will be a horizontal line.
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0
will be
138
¤
CHAPTER 2
LIMITS AND DERIVATIVES
6.
7.
8.
9.
10.
11.
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SECTION 2.8
12. The slopes of the tangent lines on the graph of
positive, so the -values of
=
=
THE DERIVATIVE AS A FUNCTION
( ) are always
0
( ) are always positive. These values start
out relatively small and keep increasing, reaching a maximum at about = 6. Then the -values of graph of
0
=
0
( ) decrease and get close to zero. The
tells us that the yeast culture grows most rapidly after 6 hours
and then the growth rate declines. 13. (a)
0
( ) is the instantaneous rate of change of percentage
of full capacity with respect to elapsed time in hours. (b) The graph of 0 ( ) tells us that the rate of change of percentage of full capacity is decreasing and approaching 0. 14. (a)
0
( ) is the instantaneous rate of change of fuel economy with respect to speed.
(b) Graphs will vary depending on estimates of will change from positive to negative at about
0
, but = 50.
(c) To save on gas, drive at the speed where is a maximum and 0 is 0, which is about 50 mi h. 15. It appears that there are horizontal tangents on the graph of
for = 1963
and = 1971. Thus, there are zeros for those values of on the graph of 0
. The derivative is negative for the years 1963 to 1971.
16. See Figure 3.3.1. 17.
The slope at 0 appears to be 1 and the slope at 1 appears to be 2 7. As
decreases, the slope gets closer to 0. Since
the graphs are so similar, we might guess that
0
( )=
.
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¤
139
140
¤
CHAPTER 2
LIMITS AND DERIVATIVES
18.
As
numbers to 1. As 0
As a guess,
0
19. (a) By zooming in, we estimate that
and
0
0
0
0
(− ) = − 0 ( ). So
0
( + )− ( )
( ) = lim
→0
( )=1
makes sense.
(1) = 2,
− 12
0
= −1,
0
(−1) = −2,
= lim
+2
+
2
−
( )=2 .
2
= lim
2
2
+
= lim
→0
0
20. (a) By zooming in, we estimate that 0
(b) By symmetry,
(− ) =
0
(0) = 0,
( ). So
( + )− ( )
( ) = lim
≈ 0 75,
0 1 2
(d) Since
0
Using
0
→0
0
2
+3
2
+
= lim →0
(1) ≈ 3, 0
0
(2) ≈ 12, and
(−2) ≈ 12, and
(1) = 3, we have
( + ) −
3
0
(−1) ≈ 3,
3
= lim
→0
3
= lim (2 + ) = 2 →0
0
(0) = 0, it appears that
3
= lim
(2 + )
→0
− 21 ≈ 0 75,
0
(c)
→0
( + )2 −
2
→0
= lim
0
→0
2
0
0
= 1,
0
or
( ) is twice the value of , so we guess that
= lim
(e)
2
( ) gets closer to 0.
(−2) = −4.
(c) It appears that (d)
0
becomes large,
( )=1
0 1 2
(0) = 0,
( ) decreases from very large
(2) = 4.
(b) By symmetry, and
0
increases toward 1,
(
3
2
0
(3) ≈ 27.
(−3) ≈ 27.
may have the form
= 3, so
+3
0
+3
0
( )=3
2
3
+
2
.
)−
3
→0
(3
2
+3
+
2
)
= lim (3 →0
2
+3
+
2
)=3
2
0
( )=
2
.
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SECTION 2.8
21.
0
( + )− ( )
( ) = lim
→0
= lim
→0
3
22.
0
0
= domain of
= R.
( + )− ( )
→0
= lim
→0
0
+ )
+ −
+
= lim
−
→0
=
→0 0
= domain of
= R.
( + )− ( )
( ) = lim
2 5( + )2 + 6( + ) − 2 5
= lim
→0
2
+6
→0 2
= lim
[ ( + )+ ]− (
= lim
→0
= lim
23.
3 +3 −8−3 +8
→0
→0
( ) = lim
Domain of
= lim
= lim 3 = 3
→0
Domain of
[3( + ) − 8] − (3 − 8)
= lim
THE DERIVATIVE AS A FUNCTION
2 5(
2
+2
2
2
−6
) +6 +6 −25
+
25
= lim
→0
2
+5
2
+6 − 25
+25
→0
2
= lim
5
+2 5
+6
(5 + 2 5 + 6)
= lim
→0
= lim (5 + 2 5 + 6) →0
→0
= 5 +6 Domain of 24.
0
0
= domain of
= R.
( + )− ( )
( ) = lim
4 + 8( + ) − 5
= lim
→0
2
)
→0 2
= lim
+ )2 − (4 + 8 − 5
4 + 8 + 8 − 5(
2
+2
2
) − 4− 8 +5
+
2
8 −5
= lim
→0
2
−
−5
→0
2
= lim
8 −1
−5
(8 − 10 − 5 )
= lim
→0
= lim (8 − 10 − 5 )
→0
→0
= 8 − 10 Domain of 25.
0
0
= domain of
= R.
( + )− ( )
( ) = lim
+ )2 − 2
= lim
→0 2
+2
= lim
3
−2
+
2
2
−6
−6
3
−2
3
)
2
−
3
+2 2
2
+
2
−6
2
−6
2
−2
= lim
→0
(2 +
−6
2
−6
→0
= lim (2 + Domain of
−2
3
→0
→0
2
→0 2
= lim
+ ) 3] − (
−6
= domain of
2 0
−6
−2
2
)=2 −6
= R. 1
1
2
√
−
√
+
−2
)
2
+5
¤
141
26.
0
( + )− ( )
( ) = lim
√ = lim
→0
√
− √ = lim
→0
√
→0
= √ √
√
+
−1 √
√
√
=
+
= domain of
+
0
√
+
√
√ = lim
→0
−( + )
= lim
Domain of
+
= lim →0
+
−1 √ 2
√
+
√
√
→0
− √
+
√
−
√
+
= lim √ +
→0
+
√ ·√
√ +
√
+ +
−1 √
√
+
√
+
√ +
+
1
=− 2
3 2
= (0 ∞).
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142
¤
CHAPTER 2
LIMITS AND DERIVATIVES
√ 27.
0
( + )− ( )
( ) = lim →0
9− ( + )+
9− ( + )+
→0
0
→0
=
√ 9−
= (−∞ 9], domain of
( + )− ( )
√ 9−
√ 2 9−
= (−∞ 9). 2
−
+ )−3
2
= lim
→0
9− ( + )+
−1
( + )2 − 1 ( ) = lim
9−
−
= lim
√ 9−
−1
= lim
0
9−( + )+
√ 9− ( + )+ 9−
[9 − ( + )] − (9 − )
→0
28.
9−
→0
= lim
Domain of
9− ( + )−
= lim
√
−1
2 −3
→0
+ )2 − 1](2 − 3) − + ) − 3]( [2( + ) − 3](2 − 3)
= lim
2
− 1)
→0
= lim
2
(
+2
2
+
− 1)(2 − 3) − (2 + 2 − 3)(
3
(2
2
2
+4
2
−2 −3
+2
→0
Domain of
29.
0
( ) = lim
4
2
2
+2
−6
−3
−2
2
2
+2
3
+ 3) − (2
2
2
+2
− 6 − 3 +2
=
(2 + 2 − 3)(2 − 3) = domain of ( + )−
0
= (−∞
()
2
2
−3
+2
− 2 − 2 + 3)
+2
− 6 − 3 + 2)
(2 + 2 − 3)(2 − 3)
→0
2
2
−6 +2
(2 − 3)2
3 2)
= lim
2
(2
= lim
(2 + 2 − 3)(2 − 3)
→0
= lim
−3
(2 + 2 − 3)(2 − 3)
→0
= lim
2
−6
→0
= lim
− 1)
+ ) − 3](2 − 3)
→0
= lim
2
∪ ( 32 ∞).
1 − 2( + ) − 1− 2 3+ ( + ) 3+
→0
[1 − 2( + )](3 + ) − [3 + ( + )](1 − 2 ) [3 + ( + )](3 + )
→0
= lim
3+
−6 −2
→0
= lim
2
−6 −2
− (3 − 6 + − 2
2
+
−2 )
→0
[3 + ( + )](3 + ) −7
= lim
−7
= lim
=
−7
−6 − (3 + + )(3 + )
→0
Domain of 30.
0
( ) = lim
= (−∞ −3) ∪ (−3 ∞). ( + )− ( ) ( + )3 2 − 3 = lim
Domain of
[( + )3 3
2
2
3
+3
( + )3
+ 2
= domain of
= 3 2
+ 0
→0
3 2]
2
2
+3
= lim
+
3 2
+ )3
= lim →0
3
( + ) −
= lim
→0
2
→0 3
= lim
(3 + )2
(3 + + )(3 + )
0
= domain of
→0
→0
→0
+ + )(3 + )
2
[( + )3
2
[
+
−
3 2
][( + )3
+ )3
−
= lim →0
3 2]
2
+
2
3
3 2
+
]
3 2]
2
3
3
+
+3
2
2
+3
[( + )3
+ 2
+
3 2]
2
= 3 2
3
1 2
2
= [0 ∞). Strictly speaking, the domain of
0
is (0 ∞) because the limit that defines
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0
(0) does
SECTION 2.8
THE DERIVATIVE AS A FUNCTION
¤
143
not exist (as a two-sided limit). But the right-hand derivative (in the sense of Exercise 64) does exist at 0, so in that sense one
31.
0
to be [0 ∞).
0
could regard the domain of
( + )− ( )
( ) = lim →0
2
4
2
= lim
+4
+6
0
+6
2 2
+4
3
+
4
−
4
4
+
= lim 4
3
→0
= domain of
3
+4
→0
3
→0
Domain of
4
4
→0 3
= lim
( + )4 −
= lim
2
+6
2
+4
+
3
=4
3
0
; but as
= R.
32. (a)
(b) Note that the third graph in part (a) has small negative values for its slope, See the graph in part (d). ( + )− ( ) (c) 0 ( ) = lim →0
6− ( + )−
= lim
(d) √ 6−
√ 6− ( + )+ 6− √ 6−( + )+ 6−
→0
[6 − ( +
= lim →0
− (6 − )
6− ( + )+
√ + 6−
−
−
√ 6−
−1
= lim √ →0 6−
→ 6− ,
=
√
= lim
→0
6−
−
+
√ 6−
−1 √ 2 6−
= (−∞ 6], domain of 0 = (−∞ 6). ( + )− ( ) + )4 + 2( + )] − ( 0 ( ) = lim = lim
Domain of 33. (a)
→0
4
+2 )
→0 4
= lim
3
2
+4
+6
3
2
2
3
+4
4
+
4
+2 +2 −
−2
→0
4
= lim
+6
2
3
+4
+
4
+2
→0
(b) Notice that
0
3
→0
= lim (4 →0
(4
= lim
3
+6
2
+4
( ) = 0 when
2
+
3
+ 2) = 4
3
+2
has a horizontal tangent,
positive when the tangents have positive slope, and negative when the tangents have negative slope.
0
0
( ) is
( ) is
+6
2
+4
2
+
3
+ 2)
0
→ −∞.
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144
¤
34. (a)
CHAPTER 2
0
LIMITS AND DERIVATIVES
( + )− ( )
( ) = lim
+ ) + 1 ( + )] − ( + 1
= lim
→0
→0
2 →0
0
−
2
+
= lim
+ 1)
→0
( + ) ( ) = 0 when
2
+2
(
= lim
2
− 1)
+
2
= lim →0
( + ) 0
has a horizontal tangent, 0
positive when the tangents have positive slope, and
−1
+
3
+ )−( ( + )
+
→0
2
(
= lim
2
+1
→0 3
2
[( + ) + 1] − ( + )( ( + )
= lim
( + )2 + 1 − +
= lim
→0 2
(b) Notice that
)
2
=
−1
+
+
, or 1 −
2
( + )
2
+ )
1 2
( ) is
( ) is
negative when the tangents have negative slope. Both functions are discontinuous at
35. (a)
= 0.
0
( ) is the rate at which the unemployment rate is changing with respect to time. Its units are percent unemployed per year.
(b) To find
0
( + )−
( ), we use lim
()
(2004) − 0
For 2003:
(2003) ≈ 0
= −1
⇒
0
(2004) ≈
(2003) −
⇒
0
(2004) ≈ 0
So we estimate that
0
36. (a)
0
()
= −1 and
1 2 [−0
= 1, and then average the two results to obtain a final estimate.
60− 55
(2004) =
−1
= −0 5;
51− 55
(2004) =
2005 − 2004
(2004) ≈
for small values of .
= −0 5
1
2003 − 2004 (2005) −
= 1
=
(2004) by using
()
55− 60
(2003)
2004 − 2003
For 2004: We estimate
( + )−
≈
→0
= −0 4.
1
5 + (−0 4)] = −0 45. Other values for
0
( ) are calculated in a similar fashion.
2003
2004
2005
2006
2007
2008
2009
2010
2011
2012
−0 50
−0 45
−0 45
−0 25
0 60
2 35
1 90
−0 20
−0 75
−0 80
( ) is the rate at which the number of minimally invasive cosmetic surgery procedures performed in the United States is
changing with respect to time. Its units are thousands of surgeries per year. (b) To fin
0
( + )−
( ), we use lim
()
→0
(2002) − 0
For 2000:
(2000) ≈
For 2002: We estima
(2000) − = −2
⇒
0
(2002) ≈
⇒
0
(2002) ≈
for small values of .
= 2, and then average the two results to obtain a final estimate.
5500 − 4897
(2002)
(2002)
= −301 5
2
= −2 and
2000 − 2002 (2004) −
= 2
=
(2002) by using
()
4897 − 5500
(2000)
2002 − 2000 0
( + )−
≈
=
−2
7470 − 4897
= −301 5
2004 −
2002 2 So we estimate tha
0
(2002) ≈
0
()
1286 5
1 2 [−301
5 + 1286 5] = 492 5.
2000
2002
2004
2006
2008
2010
2012
−301 5
492 5
1060 25
856 75
605 75
534 5
737
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SECTION 2.8
THE DERIVATIVE AS A FUNCTION
(c)
¤
145
(d) We could get more accurate values for
0
( ) by obtaining data for
more values of .
37. As in Exercise 35, we use one-sided difference quotients for the
first and last values, and average two difference quotients for all other values. 14
21
28
35
42
49
()
41
54
64
72
78
83
0
13 7
23 14
18 14
14 14
11 14
5 7
()
38. As in Exercise 35, we use one-sided difference quotients for the
first and last values, and average two difference quotients for all other values. The units for
( ) 0
39. (a)
( )
0
( ) are grams per degree (g ◦ C).
15 5
17 7
20 0
22 4
24 4
37 2
31 0
19 8
97
−9 8
−2 82
−3 87
−4 53
−6 73
−9 75
is the rate at which the percentage of the city’s electrical power produced by solar panels changes with respect to time , measured in percentage points per year.
(b) 2 years after January 1, 2000 (January 1, 2002), the percentage of electrical power produced by solar panels was increasing at a rate of 3.5 percentage points per year. is the rate at which the number of people who travel by car to another state for a vacation changes with respect to the
40.
price of gasoline. If the price of gasoline goes up, we would expect fewer people to travel, so we would expect
to be
negative. 41.
is not differentiable at
= −4, because the graph has a corner there, and at
= 0, because there is a discontinuity there.
42.
is not differentiable at
= −1, because there is a discontinuity there, and at
= 2, because the graph has a corner there.
43.
is not differentiable at
= 1, because
44.
is not differentiable at
= −2 and
is not defined there, and at
= 5, because the graph has a vertical tangent there.
= 3, because the graph has corners there, and at
= 1, because there is a discontinuity
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146
¤
CHAPTER 2
LIMITS AND DERIVATIVES
45. As we zoom in toward (−1 0), the curve appears more and more like a straight
line, so ( ) =
= −1. But no matter how much
is differentiable at
+
we zoom in toward the origin, the curve doesn’t straighten out—we can’t eliminate the sharp point (a cusp). So
is not differentiable at
= 0.
46. As we zoom in toward (0 1), the curve appears more and more like a straight 2
line, so ( ) = (
− 1)2
3
is differentiable at
= 0. But no matter how much
we zoom in toward (1 0) or (−1 0), the curve doesn’t straighten out—we can’t eliminate the sharp point (a cusp). So is not differentiable at 47. Call the curve with the positive -intercept
= 0, but =
0
= 0, so 0
. Now 00
Therefore,
and the other curve
0
(−1) is negative since 0
(1) is greater than
is increasing. Thus,
appears to be zero since 49.
= , =
0
, =
Notice that
has a maximum (horizontal tangent) at
cannot be the derivative of . Also notice that where is positive,
00
00
is below the -axis there and
is increasing. Thus,
(1) is positive since
=
and
is concave upward at
= 1.
(−1).
48. Call the curve with the smallest positive -intercept
quadrant,
= ±1.
=
and
0
=
and the other curve 0
. Now
has an inflection point at
= 0. We can immediately see that can be neither
(−1) is positive since
= 1. Therefore,
. We can see this because where
Notice that where
0
0
is positive in the first
is above the -axis there and
(1) is greater than
00
00
(1)
(−1).
has a horizontal tangent, = 0, and where has a horizontal tangent, nor
0
, since at the points where has a horizontal tangent, neither
nor is equal to 0. 50. Where
has horizontal tangents, only is 0, so
negative for
0, so
domain is , so
0
0
= .
= .
has negative tangents for
has positive tangents on R (except at
= . We conclude that
51. We can immediately see that
= , =
0
, =
00
, and
0
the velocity function, and hence,
= 0), and the only graph that is positive on the same =
000
.
0
has a horizontal tangent,
= 0 at the point where has a horizontal tangent, so must be the graph of
= . We conclude that is the graph of the position function.
must be the jerk since none of the graphs are 0 at its high and low points. where has a maximum, so
0 and is the only graph that is
is the graph of the acceleration function, since at the points where
neither nor is equal to 0. Next, we note that
52.
0
= . We conclude that
is 0 where has a maximum, so
0
= .
is the position function, is the velocity, is the acceleration, and
the jerk. 53.
0
( ) = lim
( + )− ( )
= lim
→0
= lim
[3( + )2 + 2( + ) + 1] − (3
(3
2
+6
+3
→0
+ 2 + 1)
2
+ 2 + 2 + 1) − (3
2
+ 2 + 1)
→0
= lim
2
→0 2
= lim
6
+3
+2
→0
(6 + 3 + 2)
is 0
= lim (6 + 3 + 2) = 6 + 2 →0
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
is
SECTION 2.8 0 00
( ) = lim
( + )−
0
( )
[6( + ) + 2] − (6 + 2)
= lim
→0
6
= lim
¤
(6 + 6 + 2) − (6 + 2)
= lim
→0
THE DERIVATIVE AS A FUNCTION
→0
= lim 6 = 6
→0
→0
We see from the graph that our answers are reasonable because the graph of 0
00
is that of a linear function and the graph of
is that of a constant
function.
54.
0
( + )− ( )
( ) = lim →0
3
= lim
2
(
+3
+ )] − (
3
− 3 − 3 )−(
+
2
−3 )
→0 2
− 3)
+
= lim (3
→0
→0
0 00
( ) = lim
( + )−
0
( )
= lim
= lim
→0
2
+3
+
2
[3( + )2 − 3] − (3
3) = 3
2
− 3)
2
3
+3
→0
−
2
+3
(3
= lim
−3 ) 3
→0 3
2
+3
+ )3 − 3
= lim
−3
− 2
3
2
(3
= lim
→0
3
+
2
+6
+3
2
− 3) − (3
− 3)
→0
2
= lim
6
+3
(6 + 3 )
= lim
→0
= lim (6 + 3 ) = 6 →0
→0
We see from the graph that our answers are reasonable because the graph of 0
is that of an even function ( is an odd function) and the graph of 0
that of an odd function. Furthermore, tangent and 55.
0
( ) = lim
( + )− ( )
(4 + 2 − 3
has a horizontal
has a horizontal tangent. 2
−
3
)
−3
−
2
)
= lim (4 + 2 − 3
2
−3
−
2
)=4 −3
2
→0
0
( ) = lim
= 0 when
→0 2
→0
00
= 0 when
0
2( + )2 − ( + )3 − (2
= lim
→0
= lim
00
( + )−
0
( )
4( + ) − 3
+ )2 − (4 − 3
2
(4 − 6 − 3 )
)
= lim
= lim
→0
→0
→0
= lim (4 − 6 − 3 ) = 4 − 6 →0
00 000
( ) = lim
( + )−
00
( )
= lim
→0
( ) = lim
→0
+ )] − (4 − 6 )
= lim
→0
000 (4)
[4 − 6
( + )−
000
( )
= lim
→0
−6
→0
−6 − (−6)
= lim
→0
0
= lim (0) = 0 →0
= lim ( − 6) = −6 →0
00
is
147
The graphs are consistent with the geometric interpretations of the derivatives because maximum,
00
0
has zeros where
has a zero where
0
has a local minimum and a local
has a local maximum, and
constant function equal to the slope of
00
000
is a
.
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148
¤
CHAPTER 2
LIMITS AND DERIVATIVES
56. (a) Since we estimate the velocity to be a maximum
at = 10, the acceleration is 0 at = 10.
appears to decrease by 10 ft s2 over a period of 20 s.
(b) Drawing a tangent line at = 10 on the graph of ,
So at = 10 s, the jerk is approximately −10 20 = −0 5 (ft s2 ) s or ft s3 . −
57. (a) Note that we have factored
as the difference of two cubes in the third step. 1 3 1 3 − 1 3 − 1 = lim = lim
( )− ( )
0
( ) = lim
−
→
−
→
1
= lim →
2 3
1
=
1 3 1 3
+
→
+
2 3
3
or
1 3
(
−
1 3 )( 2 3
3
1 3 1 3
+
+
2 3)
−2 3
1 3
2 3
√ 3 (b)
(0 + ) − (0)
0
(0) = lim
−0
= lim
→0
1
= lim
→0
→0
2 3
exist, and therefore 0(0) does not exist. 1 = ∞ and is continuous at (c) lim 0 ( ) = lim →0
58. (a)
→0
0
(0) = lim
3
2 3
= lim
−0
(c) ( ) =
2 3
+
+
is continuous at
1 3
(
→
1 3
(
−
lim ( ) = lim →0
→0
1 3
2 3
2 3
2
= 2 3
3
or
1 3
2
59.
( ) = − 6 =
−( − 6)
+
(d)
= ∞. This shows that
= 0.
if
−6≥6
if
−6
0
=
−6 6−
if
≥6
if
6
+
1 3
1 3 1 3
3
3
−6
1 3
)(
−1 3
= 0 and
has a vertical tangent line at
1 3
1 3 )( 2 3
2 0
−
1 3
=
1 3 1 3
1 3
= lim
−
→
= lim 2 3
has a vertical tangent at
, which does not exist.
2 3
1 3
+
→
−
2 3
1 3
1
= lim →0
= lim
−
→
−0
→0
( )− ( )
(b) 0( ) = lim
= 0 (root function), so
2 3
( ) − (0)
→0
. This function increases without bound, so the limit does not
)
+
2 3)
= 0.
So the right-hand limit is lim →6+
is lim →6−
( ) − (6) −6
= lim →6−
( ) − (6) −6 − 6 − 0 −6
= lim
− 6 − 0 −6
→6+
= lim →6−
= lim
6− −6
→6+
−6 −6
= lim 1 = 1, and the left-hand limit →6+
= lim (−1) = −1. Since these limits are not equal, →6−
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 2.8 0
( ) − (6)
(6) = lim
does not exist and
is not differentiable at 6.
−6
→6
0
However, a formula for
0
is
( )=
1
if
6
−1
if
6
0
Another way of writing the formula is 60.
( )=
−6 . − 6
( ) = [[ ]] is not continuous at any integer , so at
THE DERIVATIVE AS A FUNCTION
by the contrapositive of Theorem 4. If
is not differentiable
is not an integer, then
is constant on an open interval containing , so 0( ) = 0. Thus, 0
( ) = 0,
not an integer. 2
61. (a) ( ) =
=
−
2
if
≥0
if
0
2
(b) Since ( ) =
0
0
( ) = −2 for
So
62. (a) =
−
so ( ) =
if
≥0
if
0
+ =
Graph the line (b)
( )=
2
if
≥ 0
−2
if
0
2
if
≥0
0
if
0
= 2 for
is not differentiable at
0. At
( ) − (0) = lim →0 −0
(0) = lim
→0
0
≥ 0, we have
0
( ) = 2 for
[See Exercise 19(d).] Similarly, since ( ) = − we have
(c) From part (b), we have
for
is differentiable at 0. Thus,
2
0.
for
= 0, we have
= lim = 0 →0
is differentiable for all .
= 2 .
.
≥ 0 and graph
= 0 (the x-axis) for
0.
= 0 because the graph has a corner there, but
is differentiable at all other values; that is, is differentiable on (−∞ 0) ∪ (0 ∞). (c) ( ) =
2
if
≥0
0
if
0
⇒
0
( )=
Another way of writing the formula is 63. (a) If
0
2
if
0
0
if
0
( ) = 1 + sgn
for
= 0.
is even, then 0
( − ) = lim
(− + ) − (− )
= lim
( − )− ( )
→0
= − lim
∆ →0
Therefore,
0
= lim
[−( − )] − (− )
→0
→0
= − lim
( − )− ( )
→0
0,
[let ∆ = − ]
−
( +∆ )− ( ) = − 0( ) ∆
is odd.
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
¤
149
150
¤
CHAPTER 2
(b) If
LIMITS AND DERIVATIVES
is odd, then 0
(− + ) − (− )
( − ) = lim
= lim
− ( − )+ ( )
= lim
( +∆ )− ( ) = ∆
∆ →0
64. (a)
0
( )
is even. (4 + ) − (4)
(4) = lim
−
0
[let ∆ = − ]
−
→0
= lim 0
− (− )
( − )− ( )
= lim
→0
Therefore,
[−( −
→0
→0
= lim
(b)
→0−
→0−
−
= lim
5 − (4 + ) − 1
= −1
→0−
and 0 + (4)
(4 + ) − (4)
= lim
→0
→0+
1 − (1 − )
= lim →0+
(c) ( ) =
= lim+
1 −1 5 − (4 + )
= lim
(1 − )
0
if
5−
if 0
1 (5 − ) if At 4 we have lim →4−
→0+
1
=1
1−
≤0 4 ≥4
( ) = lim (5 − ) = 1 and lim →4−
continuous at 4. Since (5) is not defined,
→4+
→4+
= 1, so lim ( ) = 1 = (4) and
→0+
is
→4
is continuous on the
( ) = lim (5 − ) = 5 = 0 = lim
→0+
(d) From (a),
5−
is discontinuous at 5. These expressions show that
intervals (−∞ 0), (0 4), (4 5) and (5 ∞). Since lim
not exist, so
1
( ) = lim
( ), lim ( ) does
→0−
→0
is discontinuous (and therefore not differentiable) at 0. is not differentiable at 4 since
0 −(4)
=
0 +(4),
and from (c),
is not differentiable at 0 or 5.
65. These graphs are idealizations conveying the spirit of the problem. In reality, changes in speed are not instantaneous, so the
graph in (a) would not have corners and the graph in (b) would be continuous. (a)
66. (a)
(b)
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
CHAPTER 2 REVIEW
(b) The initial temperature of the water is close to room temperature because of
¤
151
(c)
the water that was in the pipes. When the water from the hot water tank starts coming out,
is large and positive as
increases to the
temperature of the water in the tank. In the next phase,
= 0 as the
water comes out at a constant, high temperature. After some time, becomes small and negative as the contents of the hot water tank are exhausted. Finally, when the hot water has run out,
is once again 0 as
the water maintains its (cold) temperature.
In the right triangle in the diagram, let ∆ be the side opposite angle
67.
and ∆
the side adjacent to angle . Then the slope of the tangent line =∆
i
∆ = tan . Note that 0
that the derivative of ( ) =
2
is
0
( )=
the curve at the point (1 1) is 2. Thus, tangent is 2; that is,
2
2
. We know (see Exercise 19)
. So the slope of the tangent to
is the angle between 0 and
2
whose
= tan−1 2 ≈ 63◦ .
Review
1. False.
Limit Law 2 applies only if the individual limits exist (these don’t).
2. False.
Limit Law 5 cannot be applied if the limit of the denominator is 0 (it is).
3. True.
Limit Law 5 applies. −9
2
4. False.
is not defined when
= 3, but
+ 3 is.
−3 2
5. True.
lim →3
6. True.
−9
= lim
−3
→3
( + 3)( − 3)
= lim ( + 3)
( − 3)
The limit doesn’t exist since ( )
→3
( ) doesn’t approach any real number as
approaches 5.
(The denominator approaches 0 and the numerator doesn’t.) 7. False.
Consider lim
→5
( − 5) sin( − 5) or lim . The first limit exists and is equal to 5. By Example 2.2.3, we know that →5 −5 −5
the latter limit exists (and it is equal to 1). 8. False.
If ( ) = 1
, ( ) = −1
, and
= 0, then lim ( ) does not exist, lim ( ) does not exist, but →0
→0
lim [ ( ) + ( )] = lim 0 = 0 exists. →0
9. True.
→0
Suppose that lim [ ( ) + ( )] exists. Now lim ( ) exists and lim ( ) does not exist, but →
→
→
lim ( ) = lim {[ ( ) + ( )] − ( )} = lim [ ( ) + ( )] − lim ( ) [by Limit Law 2], which exists, and →
→
→
we have a contradiction. Thus, lim [ ( ) + ( )] does not exist. →
→
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
152
¤
10. False.
CHAPTER 2 LIMITS AND DERIVATIVES
1
Consider lim [ ( ) ( )] = lim ( − 6) →6
. It exists (its value is 1) but (6) = 0 and (6) does not exist,
−6
→6
so (6) (6) = 1. 11. True. 12. False.
A polynomial is continuous everywhere, so lim ( ) exists and is equal to ( ). →
Consider lim [ ( ) − ( )] = lim
1
→0
2
→0
1
−
. This limit is −∞ (not 0), but each of the individual functions
4
approaches ∞. 13. True.
See Figure 2.6.8.
14. False.
Consider ( ) = sin
15. False.
Consider ( ) =
16. False.
The function ( )=
≥ 0. lim
for
( ) = ±∞ and
→∞
1 ( − 1)
if
=1
2
if
=1
must be continuous in order to use the Intermediate Value Theorem. For example, let
1
if 0 ≤
−1
if
3
There is no number ∈ [0 3] with ( ) = 0.
=3
17. True.
Use Theorem 2.5.8 with
18. True.
Use the Intermediate Value Theorem with
= 2, = 5, and ( ) = 4
19. True, by the definition of a limit with
For example, let ( ) = Then ( )
21. False. 22. True.
= −1,
2
− 11. Note that (4) = 3 is not needed.
= 1, and
= , since 3
4.
= 1. 2
20. False.
has no horizontal asymptote.
+ 1 if
=0
if
=0
2
1 for all , but lim ( ) = lim →0
→0
2
+ 1 = 1.
See the note after Theorem 2.8.4. ( ) exists ⇒
0
is differentiable at
⇒
is continuous at
⇒
lim ( ) = ( ). →
2
2
is the second derivative while
23. False.
is the first derivative squared. For example, if
= ,
2 2
2
= 0, but
then
= 1.
2
24. True.
( )= −4
0
10
− 10
2
+ 5 is continuous on the interval [0 2], (0) = 5, (1) = −4, and (2) = 989. Since
5, there is a number in (0 1) such that ( ) = 0 by the Intermediate Value Theorem. Thus, there is a
root of the equation
10
25. True.
See Exercise 2.5.72(b).
26. False
See Exercise 2.5.72(b).
− 10
2
+ 5 = 0 in the interval (0 1). Similarly, there is a root in (1 2).
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
CHAPTER 2 REVIEW
1. (a) (i) lim+ ( ) = 3
(ii)
→2
(iii) lim
→−3
lim
¤
( )=0
→−3+
( ) does not exist since the left and right limits are not equal. (The left limit is −2.)
(iv) lim ( ) = 2 →4
(v) lim ( ) = ∞ →0
→2−
(vii) lim
( )= 4
( ) = −1
(viii) lim
→∞
→−∞
= −1 and
(b) The equations of the horizontal asymptotes are (c) The equations of the vertical asymptotes are
2.
( ) = −∞
(vi) lim
= 0 and
= 4.
= 2.
(d) is discontinuous at = −3, 0, 2, and 4. The discontinuities are jump, infinite, infinite, and removable, respectively. lim ( ) = −2, lim ( ) = 0, lim ( ) = ∞, →−∞
lim
→3−
→∞
( ) = −∞,
→−3
lim+ ( ) = 2, →3
is continuous from the right at 3
3−
3. Since the exponential function is continuous, lim
→1 2
4. Since rational functions are continuous, lim →3
2
5. lim →−3
2
+2 −3
2
6. lim →1+
7. lim
2
−9
−9
= lim →−3
( + 3)( − 3)
2
2
−9
1−1
0
=
3 −9
=
−3
= lim
=
−1
→−3
+ 2 − 3 → 0+ as
→0
=
32 + 2(3) − 3 −3 − 3
3
−3
2
+3 −1 +1
lim =
→0
3
→0
−3
2
0
= 0.
12 −6
=
−4
2
→ 1+ and 2
= lim
=
−3 − 1
+2 −3
( − 1)3 + 1
= 1.
2
+2 − 3
( + 3)( − 1)
= −∞ since
=
3 2
−9
0 for 1
+2 − 3
+3 = lim →0
2
−3 +3 =3
Another solution: Factor the numerator as a sum of two cubes and then simplify. lim →0
( − 1)3 + 1
= lim
( − 1)3 + 13
→0
= lim ( − 1)2 − 0
= lim
[
− 1) + 1] ( − 1)2 − 1( − 1) + 12
→0
+2 = 1 − 0+ 2 = 3
3.
153
→ 2
8. lim →2
3
−4
= lim
−8
→2
( + 2)( − 2) ( − 2)(
2
+2
= lim
+ 2 + 4)
→2
2
+2 +4
√ 9. lim →9
( − 9)4
2+ 2
=
=
4
4+4+4
12
0 for
= 9.
=
1 3
√ = ∞ since ( − 9) → 0+ as 4
→ 9 and
( − 9)4
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
154
¤
CHAPTER 2 LIMITS AND DERIVATIVES
4−
10. lim →4+
4−
= lim
4 −
→4+
1
= lim
−(4 − )
→4+
= −1
−1 2
4
11. lim →1
3
+5
√
2
= lim
−6
3
−3
2
(
→1
(
√
+6 −
12. lim →3
−1
→3
2
+ 1)( + 1)( − 1)
(
= lim
+6+
→3
( 2(
2
→3
→3
√ is positive,
2
2
√
→∞
√
2
2
→
15. Let = sin . Then as
→−∞
5+
−
4
−3
4
2
=
−9
√
−9
2
−
√ →−∞
17. lim →∞
2
+4 +1 −
2
+6 + − 6)
√
−( − 3)( + 2)
= lim →3
+6 +
− 3)
2(
√
+6+
−
54
√
−9
1
2
= lim →∞
√
2
9
=
2− 6
1− 0
=
2− 0
1 2
→−∞
√
2
1
= lim
(2 − 6) (− )
→−∞
2
−
−3
2
= lim
4
)
4)
4
= 4
lim →−∞
√ +4 +1 −
=
−2 − +6
√ 1− 0
=−
−2 + 0
1 2
1
→0+
1
4
5
4
−2
2
+1
3
√
−1
=
−3
0− 0− 1
2
2
= lim →∞
+4 +1+
(4 + 1)
= lim (
√
2
( √
divide by
=
4+0
=
√
=
for
4
=2
−1 −3
1
=
3
+ 4 + 1) − 2
+4 +1 +
2
+4 +1+ )
4+1
=
0+ 0− 3
2
+4 +1+ ·
→∞
= lim
2
9
→
(5 +
→∞
−9
→ 0+ , so → 0+ . Thus, lim− ln(sin ) = lim ln = −∞.
(1 − 2
lim
2
= lim
, sin
=
1(7)
5
=−
(2 − 6)
→∞
√ √
2
= lim
2 −6
→−∞
1− 2
( + 6)
2(2)
= = − . Thus, √
lim
−
− 3)
2(
9(3 + 3)
2 −6
lim
16.
+ 6 )2 − √
=
= = . Thus, lim
is negative,
5
=−
+6+
√
14. Since
→3
+6+
−( + 2)
= lim
13. Since
− 3)
2(
+ 1)( + 1)
→1
− 3)
−(
= lim
√
√
2
(
2
+6 −
= lim
= lim
( + 6)( − 1)
→1
√ +6+ ·√
− 3)
2(
= lim
+ 5 − 6)
2
+6 −
= lim
− 1)
2
+ 1)(
0
2
4 7
→∞
−
18. Let =
19. Let = 1
2
1+4
1 − ). Then as
=
. Then as
+1
2
+1
√ 1+ 0+0+1
→ ∞, → −∞, and lim
→ 0+ , → ∞ , and lim tan−1 (1
→1
1 −1
1
+ 2
−3 +2
= lim →1
= lim →1
1 −1
= lim
) = lim tan−1 = →∞
1
+
( − 1)( − 2)
= lim →1
→1
1 −2
. 2 −2
= lim
( − 1)( − 2) −1
= 0.
→−∞
→∞
→0+
20. lim
− 2
2
=
( − 1)( − 2) 1
+
1 ( − 1)( − 2)
= −1
1− 2
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CHAPTER 2 REVIEW
= cos2
21. From the graph of
2
cos
2
1
2
= 0 is the horizontal ⇒
2
2
≤
, it appears that
= 0 is the vertical asymptote. Now 0 ≤ (cos ) ≤ 1
asymptote and 0
2
≤
cos ⇒
2
0≤
¤
1 ≤
2
2
. But lim 0 = 0 and →±∞ 2
lim →±∞
Thus,
1
cos
= 0, so by the Squeeze Theorem, lim →±∞
2
= 0 is the horizontal asymptote. lim
2
cos2
→0
= 0.
= ∞ because cos2
→ 1 and
2
→ 0+ as
→ 0, so
= 0 is the
2
vertical asymptote.
22. From the graph of
= ( )=
√
2
+1 −
+
√
2
− , it appears that there are 2 horizontal asymptotes and possibly 2
vertical asymptotes. To obtain a different form for , let’s multiply and divide it by its conjugate. √ √ 2 + 2 − +1+ ( 2 + + 1) − ( 2 − ) √ √ 2 2 √ =√ √ √ + +1 − − 1( ) = 2
+
2
−
2
+1+
−
2
+
+1+
2
−
2 +1
= √ 2
+
+1 +
√
−
2
Now lim →∞
1(
√ ) = lim →∞
2
+
2 +1 √ +1+
2 + (1
= lim →∞
1 + (1
)
2)
) + (1
+
[since 1 − (1
√
2
=
for
0]
)
2 = = 1, 1+1
so with
= 1 is a horizontal asymptote. For
0, we have
√
2
= = − , so when we divide the denominator by ,
0, we get √
2
+
+1+
√
2
√ =−
−
2
+
+1 + √
√
2
−
2
=−
1+
1
+
1
+
2
1−
1
Therefore, lim
1(
) = lim
→−∞
→−∞
=
2 −(1 + 1)
so
2 +1
√
= −1 is a horizontal asymptote.
2
+
= −1
+1+
√
2 + (1
= lim 2
−
→∞
−
1 + (1
) + (1
) 2)
+
1 − (1
)
155
The domain of
is (−∞ 0] ∪ [1 ∞). As
= 0 is not a vertical asymptote. As
→ 0− ,
( ) → 1, so √ → 1+ , ( ) → 3, so = 1
is not a vertical asymptote and hence there are no vertical asymptotes.
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156
¤
CHAPTER 2 LIMITS AND DERIVATIVES
23. Since 2 − 1 ≤
24. Let ( ) = −
( )≤
2
2
, ( )=
2
( ) ≤ ( ) ≤ ( ) for
25. Given
√ 3 = 3
− 2
2
− 2
3
=
−1
√ 3
28. Given
lim+ 2
2
4 √
⇒
√ √ . Now 3 − 0 = 3
√ √ 3 − 0 = 3 =
⇒
3
− 3 − (−2)
2
, then
− 1
2. Now let
−4
0 such that if 0
. So if we choose
√
(i) lim
−
if
3
√ 3

⇒ 3
= .
. First, note that if − 2
= min { 2 1}. Then 0
− 2
1, then ⇒
( 2)(2) = .
=4
2
, then 2
√
−4
, then 0
0, ( ) = 3 −
if 0 ≤
−4 ⇒
2
. This is true ⇔ √
3, ( ) = ( − 3)2 if
( ) = lim (3 − ) = 3
→0+
(ii) lim
−4
√
−4
2
⇔
. So by the definition of a limit,
( ) = lim →0−
→3−
(v) lim+ ( ) = lim+ ( − 3)2 = 0 →0
−
=0
→3−
(vi) Because of (iv) and (v), lim ( ) = 0. →3
→3
is discontinuous at 0 since lim
√
( ) = lim (3 − ) = 0
(iv) lim
→0
→3
3.
→0−
→0+
(iii) Because of (i) and (ii), lim ( ) does not exist.
(b)
− 2
5, then 0
− 4 = ∞.
→4
29. (a) ( ) =
=
⇔
− 3 ) = −2 by the definition of a limit.
0, we need
−4
. But (14 − 5 ) − 4
= 0.
− 2
⇒
2
√ , then 3 − 0
− 3 − (−2) = ( − 2)( − 1) = − 2 − 1 2
→0
5. So if we choose
− 0 =
. Then 0
0 so that if 0
1, so 0
→2
= 0, we have
lim ( ) = 0 by the Squeeze Theorem.
, then (14 − 5 ) − 4
− 0
0 so that if 0
. So take
0, we need
Thus, lim (
⇒
≤ 1 for
2
→2
3
−2
→1
. Thus, lim (14 − 5 ) = 4 by the definition of a limit.
→0
−1
, we have lim ( ) = 1 by the Squeeze Theorem.
. Then since cos 1
→0
⇔
Therefore, by the definition of a limit, lim 27. Given
2
and ( ) =
−5 − 2
2
→1
→0
0 we must find
26. Given
2
cos 1
0 such that if 0
⇔
(14 − 5 ) − 4
→1
= 0, and so lim ( ) = lim ( ) = 0
0, we need
−5 + 10
3 and lim (2 − 1) = 1 = lim
for 0
( ) does not exist.
(c)
is discontinuous at 3 since (3) does not exist.
30. (a) ( ) = 2 −
2
Therefore, lim
→2−
if 0 ≤
≤ 2, ( ) = 2 −
( ) = lim
→2−
2 −
2
if 2
≤ 3, ( ) =
− 4 if 3
4, ( ) =
if
≥ 4.
( ) = 0 = (2), = 0 and lim+ ( ) = lim+ (2 − ) = 0. Thus, lim →2 →2
→2
so is continuous at 2. lim
→3−
( ) = lim (2 − ) = −1 and lim →3−
→3+
( ) = lim+ ( − 4) = −1. Thus, →3
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CHAPTER 2 REVIEW
lim ( ) = −1 = (3), so →3
→4−
→4−
( ) = lim+ →4
→4+
157
(b)
is continuous at 3.
( ) = lim ( − 4) = 0 and lim
lim
¤
= .
Thus, lim ( ) does not exist, so is discontinuous at 4. But →4
lim+ ( ) =
= (4), so
→4
31. sin
and
Lastly, 32.
2
is continuous from the right at 4.
are continuous on R by Theorem 2.5.7. Since
is continuous on R since it’s a polynomial and the product
− 9 is continuous on R since it is a polynomial and
√
is continuous on R,
is continuous on R by Theorem 2.5.9.
is continuous on its domain R by Theorem 2.5.4.
is continuous on [0 ∞) by Theorem 2.5.7, so the composition
− 9 ≥ 0 = (−∞ −3] ∪ [3 ∞) by Theorem 2.5.9. Note that 2 − 2 = 0 on this set and √ 2 − 9 is continuous on its domain, (−∞ −3] ∪ [3 ∞) by Theorem 2.5.4. so the quotient function ( ) = 2 − 2 33.
2
− 9 is continuous on
√
sin
sin
( )=
5
−
3

2
+ 3 − 5 is continuous on the interval [1 2], (1) = −2, and (2) = 25. Since −2
0
25, there is a
number in (1 2) such that ( ) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation 5
34.
−
3
+ 3 − 5 = 0 in the interval (1 2).
( ) = cos
√
−
+ 2 is continuous on the interval [0 1], (0) = 2, and (1) ≈ −0 2. Since −0 2
0
2, there is a
number in (0 1) such that ( ) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation √ √ cos − + 2 = 0, or cos = − 2, in the interval (0 1). 35. (a) The slope of the tangent line at (2 1) is
( ) − (2)
lim →2
= lim
−2
9 −2
2
−1
= lim
−2
→2
8− 2
2
−2
→2
= lim
−2(
2
− 4)
= lim
−2
→2
−2( − 2)( + 2) −2
→2
= lim [−2( + 2)] = −2 · 4 = −8 →2
(b) An equation of this tangent line is
− 1 = −8( − 2) or
= −8 + 17.
36. For a general point with -coordinate , we have
= lim
2 (1 − 3 ) − 2 (1 − 3 ) −
→
6
= lim →
For
= 0,
= lim →
(1 − 3 )(1 − 3 )( − )
(1 − 3 )(1 − 3 )( − )
(1 − 3 )2
= 6 and (0) = 2, so an equation of the tangent line is − 2 = 6( − 0) or
and (−1) = 12 , so an equation of the tangent line is − 37. (a)
→
= ( ) = 1+2 +
2
=
1 2
= 38( + 1) or
=
3 8
4. The average velocity over the time interval [1 1 +
(1 + ) − 1
=
= 6 + 2 For
= −1,
=
3 8
+ 78. is
1 + 2(1 + ) + (1 + )2 4 − 13 4
(1 + ) − (1) ave
− )
6
= lim
6
=
(1 − 3 )(1 − 3 )
2(1 − 3 ) − 2(1 − 3 )
10 + =
4
2
10 + =
4 [continued]
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158
¤
CHAPTER 2 LIMITS AND DERIVATIVES
So for the following intervals the average velocities are: (i) [1 3]:
= 2,
(iii) [1 1 5]:
ave
= 0 5,
= (10 + 2) 4 = 3 m s ave
(ii) [1 2]:
= (10 + 0 5) 4 = 2 625 m s
(iv) [1 1 1]:
(1 + ) − (1)
(b) When = 1, the instantaneous velocity is lim
→0
∆
= ∆ ∆
is
(b) Since
800
(250) − =
(200) =
250
800
−
200
∆
= (10 + 0 1) 4 = 2 525 m s
10 = 2 5 m s. 4
= 250 − 200 = 50 in3 , and sin
=8
,
2
→0
→0
( ) − (2)
(2) = lim
−800
=−
= lim
= lim
( − 2)(
− 4 = 10( − 2) or 6
and
→0
800 [ − (
+ )]
+ )
2
.
−2 −4
(c)
−2
2
+ 2 + 2) −2
→2
40. 26 = 64, so ( ) =
→2
= lim
800
+ )
3
−2
→2
41. (a)
=
ave
= 3 2 − 4 = −0 8 lb in . So the average rate of change
which is inversely proportional to the square of
0
10 + 4
, the instantaneous rate of change of with respect t is ( + )− ( ) 800 ( + ) − 800 ∆ lim = lim = lim
→0
(b)
= 0 1,
= (10 + 1) 4 = 2 75 m s
= 800
= lim
0
ave
−0 8 lb in2 . = −0 016 50 in3
→0
39. (a)
= lim
→0
increases from 200 in3 to 250 in3 , we have ∆
38. (a) When
= 1,
= lim (
2
+ 2 + 2) = 10
→2
= 10 − 16 = 2.
( ) is the rate at which the total cost changes with respect to the interest rate. Its units are dollars (percent per year).
(b) The total cost of paying off the loan is increasing by $1200 (percent per year) as the interest rate reaches 10%. So if the interest rate goes up from 10% to 11%, the cost goes up approximately $1200. (c) As increases
increases. So
0
( ) will always be positive.
42.
43.
44.
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CHAPTER 2 REVIEW
√ 45. (a)
0
( + )− ( )
( ) = lim
3− 5
= lim
→0
+ )−
3− 5
[3 − 5
→0
− (3 − 5 )
+
3− 5
+ )+
−5
= lim
√ 3−5
−∞ ⇒
Domain of −∞
=
−5 2 3− 5
3− 5
⇒
5
∈ 0
√
3 − 5( + ) +
→0
3− 5
√ + )+ √ 3− 5
3− 5
(b) Domain of : (the radicand must be nonnegative) 3 − 5 ≥ 0
5 ≤3
3 − 5( + ) +
→0
= lim
¤
√
3
: exclude
3 5
because it makes the denominator zero;
5
∈ (c) Our answer to part (a) is reasonable because 3
0
( ) is always negative and
is always decreasing. 46. (a) As
→ ±∞, ( ) = (4 − ) (3 + ) → −1, so there is a horizontal = −1. As
asymptote at
( ) → −∞. Thus, there is a vertical asymptote at
those intervals. As 0
(c)
0
= −3.
is decreasing on (−∞ −3) and (−3 ∞), so
(b) Note that
→ −3− ,
→ −3+ , ( ) → ∞, and as
→ ±∞,
0
0
is negative on
→ −3− and as
→ 0. As
→ −3+ ,
→ −∞.
( + )− ( )
( ) = lim →0
= lim
4− ( + ) − 4− 3+ 3+ ( + )
= lim →0
(12 − 3 − 3 + 4 −
→0
(3 + ) [4 − ( + )] − (4 − ) [3 + ( + )]
→0
2
−
) − (12 + 4 + 4 − 3 −
[3 + ( + )] (3 + ) 2
−
)
[3 + ( + )](3 + ) −7
= lim →0
= lim
[3 + ( +
−7
= lim (3 + )
→0
7
=−
[3 + ( + )] (3 + )
(3 + )2
(d) The graphing device confirms our graph in part (b). 47.
is not differentiable: at continuous, and at
= −4 because
= 5 because
is not continuous, at
has a vertical tangent.
= −1 because
has a corner, at
= 2 because
is not
159
48. The graph of
has tangent lines with positive slope for
0 and negative slope for
0, and the values of fit this pattern,
so must be the graph of the derivative of the function for . The graph of has horizontal tangent lines to the left and right of the -axis and has zeros at these points. Hence, is the graph of the derivative of the function for . Therefore, of , is the graph of
0
, and is the graph of
00
.
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is the graph
160
¤
CHAPTER 2 LIMITS AND DERIVATIVES
49. Domain: (−∞ 0) ∪ (0 ∞); lim− ( ) = 1; lim+ ( ) = 0; →0
0
( )
50. (a)
0 for all
0
→0
0
in the domain; lim
→−∞
0
( ) = 0; lim
→∞
( )=1
( ) is the rate at which the percentage of Americans under the age of 18 is changing with respect to time. Its units are
percent per year (% yr). (b) To find
0
( + )−
( ), we use lim
()
(1960) − 0
For 1950:
(1950) ≈
=
for small values of .
= 0 46
10
= −10 and
(1960) by using
()
35 7 − 31 1
(1950)
1960 − 1950 0
For 1960: We estimate
( + )−
≈
→0
= 10, and then average the two results to obtain a
final estimate. (1950) − = −10
⇒
0
(1960) ≈
⇒
0
(1960) ≈
So we estimate that
0
()
1 2 [0
= 0 46
−10
34 0 − 35 7
(1960)
1970 − 1960
(1960) ≈
0
=
1950 − 1960 (1970) −
= 10
31 1 − 35 7
(1960)
=
= −0 17
10
46 + (−0 17)] = 0 145.
1950
1960
1970
1980
1990
2000
2010
0 460
0 145
−0 385
−0 415
−0 115
−0 085
−0 170
(c)
(d) We could get more accurate values for
0
( ) by obtaining data for the mid-decade years 1955, 1965, 1975, 1985, 1995, and
2005. 51.
0
( ) is the rate at which the number of US $20 bills in circulation is changing with respect to time. Its units are billions of
bills per year. We use a symmetric difference quotient to estimate 0
(2000) ≈
(2005) −
(2000).
5 77 − 4 21
(1995)
2005 − 1995
0
=
10
= 0 156 billions of bills per year (or 156 million bills per year).
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CHAPTER 2 REVIEW
52. (a) Drawing slope triangles, we obtain the following estimates:
and
0
(1987) ≈
02 10
0
(1950) ≈
11 10
= 0 11,
0
(1965) ≈
−1 6 10
¤
= −0 16,
= 0 02.
(b) The rate of change of the average number of children born to each woman was increasing by 0 11 in 1950, decreasing by 0 16 in 1965, and increasing by 0 02 in 1987. (c) There are many possible reasons: • In the baby-boom era (post-WWII), there was optimism about the economy and family size was rising. • In the baby-bust era, there was less economic optimism, and it was considered less socially responsible to have a large family. • In the baby-boomlet era, there was increased economic optimism and a return to more conservative attitudes. 53. ( ) ≤ ( )
⇔
− ( ) ≤ ( ) ≤ ( ) and lim ( ) = 0 = lim − ( ). →
→
Thus, by the Squeeze Theorem, lim ( ) = 0. →
54. (a) Note that
is an even function since ( ) = (− ). Now for any integer ,
[[ ]] + [− ]] =
−
= 0, and for any real number
which is not an integer,
[[ ]] + [− ]] = [ ]] + (− [[ ]] − 1) = −1. So lim ( ) exists (and is equal to −1) →
for all values of . (b)
is discontinuous at all integers.
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161
162
¤
CHAPTER 2 LIMITS AND DERIVATIVES
° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
2
Limits and Derivatives
2.1
The Tangent and Velocity Problems
SUGGESTED TIME AND EMPHASIS 1 2 –1
class
Essential material
POINTS TO STRESS 1. The tangent line viewed as the limit of secant lines. 2. The concepts of average versus instantaneous velocity, described numerically, visually, and in physical terms. 3. The tangent line as the line obtained by “zooming in” on a smooth function; local linearity. 4. Approximating the slope of the tangent line using slopes of secant lines. QUIZ QUESTIONS TEXT QUESTION Geometrically, what is “the line tangent to a curve” at a particular point? ANSWER There are different correct ones. Examples include the best linear approximation to a curve at a point, or the result of repeated “zooming in” on a curve. DRILL QUESTION Draw the line tangent to the following curve at each of the indicated points: y
y=f(x)
x
ANSWER y
y=f(x)
x
MATERIALS FOR LECTURE Point out that if a car is driving along a curve, the headlights will point along the direction of the tangent line. Discuss the phrase “instantaneous velocity.” Ask the class for a definition, such as, “It is the limit of average velocities.” Use this discussion to shape a more precise definition of a limit. Illustrate that many functions such as x 2 and x 2 sin x look locally linear, and discuss the relationship of this property to the concept of the tangent line. Then pose the question, “What does a secant line to a linear function look like?” 49
CHAPTER 2 LIMITS AND DERIVATIVES 3 Show that the slopes of the tangent lines to f x x and g x x are not defined at x 0. Note that f has a tangent line (which is vertical), but g does not (it has a cusp). The absolute value function can be explored graphically.
WORKSHOP/DISCUSSION
Estimate slopes from discrete data, as in Exercises 2 and 7.
Estimate the slope of y
3 1
x2
3
at the point 1
using the graph, and then numerically. Draw the
2
tangent line to this curve at the indicated point. Do the same for the points 0 3 and 2 ANSWER 1 5, 0,
3 5
.
48
Draw tangent lines to the curve y
sin
1 x
1 and x 2π
at x
1 . Notice the difference in the π 2
quality of the tangent line approximations.
GROUP WORK 1: WHAT’S THE PATTERN? The students will not be able to do Problem 3 from the graph alone, although some will try. After a majority of them are working on Problem 3, announce that they can do this numerically. If they are unable to get Problem 6, have them repeat Problem 4 for x 15, and again for x 0. ANSWERS 1, 2.
y
3. 2
3
0 268,
5
3
45
35
0 250,
2
0 236, 42 04
2
4.
1 4
is a good estimate.
5.
1
is a good estimate. 1
1
50
6
38
0 250
_1
6.
0
1
2
3
4
5
6
7
8
x
52
2 a
1
SECTION 2.1 THE TANGENT AND VELOCITY PROBLEMS
GROUP WORK 2: SLOPE PATTERNS When introducing this activity, it may be best to fill out the first line of the table with your students, or to estimate the slope at x 1. If a group finishes early, have them try to justify the observations made in the last part of Problem 2. ANSWERS 1. (a) 0, 0 2, 0 4, 0 6
(b) 11 5
2. (a) Estimating from the graph gives that the function is increasing for x 3 2 x 3 2, and increasing for x 3 2.
2, decreasing for
(b) The slope of the tangent line is positive when the function is increasing, and the slope of the tangent line is negative when the function is decreasing. (c) The slope of the tangent line is zero somewhere between x 2 and 3 1, and somewhere between x 3 1 and 3 2. The graph has a local maximum at the first point and a local minimum at the second. (d) The tangent line approximates the curve worst at the maximum and the minimum. It approximates best at x 0, where the curve is “straightest,” that is, at the point of inflection. HOMEWORK PROBLEMS CORE EXERCISES 1, 5, 9 SAMPLE ASSIGNMENT 1, 3, 5, 9 EXERCISE 1 3 5 9
D
A
51
N
G
GROUP WORK 1, SECTION 2.1 What’s the Pattern? Consider the function f x
1
x.
1. Carefully sketch a graph of this function on the grid below.
y 3 2 1 _1
0
1
2
3
4
5
6
2. Sketch the secant line to f between the points with x -coordinates x
7
2 and x
8
x 4.
3. Sketch the secant lines to f between the pairs of points with the following x -coordinates, and compute their slopes: (a) x 2 and x 3 (b) x 3 and x 4 (c) x 2 5 and x 3 5 (d) x 2 8 and x 3 2
52
What’s the Pattern?
4. Using the slopes you’ve found so far, estimate the slope of the tangent line at x
5. Repeat Problem 4 for x
3.
8.
6. Based on Problems 4 and 5, guess the slope of the tangent line at any point x
53
a, for a
1.
GROUP WORK 2, SECTION 2.1 Slope Patterns 1. (a) Estimate the slope of the line tangent to the curve y information to fill in the following table: x
0 1x 2 , where x
0, 1, 2, 3. Use your
slope of tangent line
0 1 2 3 (b) You should notice a pattern in the above table. Using this pattern, estimate the slope of the line tangent to y 0 1x 2 at the point x 57 5.
2. Consider the function f x
0 1x 3
3x .
(a) On what intervals is this function increasing? On what intervals is it decreasing?
(b) On what interval or intervals is the slope of the tangent line positive? On what interval or intervals is the slope of the tangent line negative? What is the connection between these questions and part (a)?
(c) Where does the slope of the tangent line appear to be zero? What properties of the graph occur at these points?
(d) Where does the tangent line appear to approximate the curve the best? The worst? What properties of the graph seem to make it so?
54
2.2
The Limit of a Function
SUGGESTED TIME AND EMPHASIS 1 class
Essential material
POINTS TO STRESS 1. The various meanings of “limit” (descriptive, numeric, graphic), both finite and infinite. Note that algebraic manipulations are not yet emphasized. 2. The geometric and limit definitions of vertical asymptotes. 3. The advantages and disadvantages of using a calculator to compute a limit. QUIZ QUESTIONS TEXT QUESTION What is the difference between the statements “ f a
L ” and “ lim f x x
L ”?
a
ANSWER The first is a statement about the value of f at the point x a the second is a statement about the values of f at points near, but not equal to, x a. DRILL QUESTION The graph of a function f is shown below. Are the following statements about f true or false? Why? (a) x a is in the domain of f (b) lim f x exists (c) lim f x is equal to lim f x x
a
x
a
x
a
y
a
0
x
ANSWER (a) True, because f is defined at x
a.
(b) True, because as x gets close to a, f x approaches a value. (c) True, because the same value is approached from both directions. MATERIALS FOR LECTURE Present the “motivational definition of limit”: We say that lim f x x
a
L if as x gets close to a, f x gets
close to L , and then lead into the definition in the text. (Note that limits will be defined more precisely in Section 2.4.) Describe asymptotes verbally, and then give graphic and limit definitions. If foreshadowing horizontal asymptotes, note that a function can cross a horizontal asymptote. Perhaps foreshadow the notion of slant asymptotes, which are covered later in the text. Discuss how we can rephrase the last section’s concept as “the slope of the tangent line is the limit of the slopes of the secant lines as x 0”. Stress that if lim f x and lim g x , then we still don’t know anything about x
a
x
a
55
lim f x
x
a
g x .
57
CHAPTER 2 LIMITS AND DERIVATIVES
WORKSHOP/DISCUSSION Explore the greatest integer function f x hand limits. Explore lim 2 x
0
21
x
[[x ]] on the interval [ 1 2] in terms of left-hand and right-
, looking carefully at 21
x
from both sides for small x . Discuss in graphical,
numerical, and algebraic (“what happens to 1 x when x is small?”) contexts. ANSWER The left-hand limit is 2, because 21 x vanishes. The right-hand limit is . Discuss a limit such as lim x
3
x
3
4x
2
. When can you “just plug in the numbers”?
GROUP WORK 1: AN INTERESTING FUNCTION Introduce this activity with a review of the concepts of left- and right-hand limits. Also make sure that the students can articulate that when a denominator gets small, the function gets large, and vice versa. (This will be needed for the second question in Problems 2 and 3.) When a group is done, inform them that one of them will be chosen at random to discuss the answer with the class, so all should be able to describe their results. 1 When they graph y , they are expected to use graphing technology. When they are all finished, 1 21 x have a different person present the solution to each part. ANSWERS 1. 2. The limit is 0. When x is small and positive, 1 x is large y and positive, so 1 21 x is large and negative. Therefore its reciprocal is very small and negative, approaching zero. 1 0
3. The limit is 1. When x is small and negative, 1 x is large 1
x
and negative, and 1 21 x is very close to 1. 4. The limit doesn’t exist because the left- and right-hand limits are not equal.
GROUP WORK 2: INFINITE LIMITS After the students are finished, Problem 2 can be used to initiate a discussion of left and right hand limits, and of the precise definition of a vertical asymptote, as presented in the text. In addition, Section 2.6 can be foreshadowed by asking the students to explore the behavior of
56
3x 2 4x 5 for large positive and large 16x 4 81
negative values of x , both on the graph and numerically. If there is time, the students can be asked to analyze the asymptotes of f x sec x and the other trigonometric functions.
58
SECTION 2.2 THE LIMIT OF A FUNCTION
ANSWERS 1. Answers will vary. The main thing to check is that there are vertical asymptotes at 2.
There are vertical asymptotes at x
y
2
and at
2.
5.
20
10
_ 4
_2
2
0
4
x
GROUP WORK 3: THE SHAPE OF THINGS TO COME This activity foreshadows concepts that will be discussed later, but can be introduced now. The idea is to show the students that the concept of “limit” can get fairly subtle, and that care is needed. The second page anticipates Section 2.6, and the third page anticipates Section 2.4. Pages 2 and 3 are independent of each other; either or both can be used. Problem 4 on page 3 is a little tricky and can be omitted if desired. ANSWERS PAGE 1 1. 21
2, 0 6
2.
2
0 36, 0 8
3
0 512, 04
0, 1 01 3. fn 1
y f§ f£ fª
2
fÁ
fö
0
_1
1
1 0829 1 for all n.
4. The curves all go through the origin.
1
_1
8
2 x
57
CHAPTER 2 LIMITS AND DERIVATIVES
PAGE 2
1. (a) 0
(b) 1
(c) 1
2. (a) 0
(b) 0
(c) 0
(d) 1
3. The function g x is important in real analysis. Its graph looks like this: y 1
1 x
0
4. (a) 1
(b) 0
PAGE 3
1.
3.
1 10
(or any positive number less than
101 10
1
1 10 )
2. Estimates will vary.
0 0049876 (or any positive number less than 0 0049876)
4. Yes, the problems could have been done with any smaller positive number. 5. The students can be forgiven for not answering this question. It will be fully answered in Section 2.4. The short answer: Let a be the “small number you can name.” Then we have shown that we can always find a small interval about x such that x 2 0 a. A similar argument can be made for the second part. The main idea here is to set up ideas that will be explored more fully in Section 2.4.
GROUP WORK 4: WHY CAN’T WE JUST TRUST THE TABLE? This activity was inspired by the article “An Introduction to Limits” from College Mathematics Journal, January 1997, page 51, and extends Example 4. Put the students into groups and give each group two different digits between 1 and 9, and then let them proceed with the problems in the handout. 58
SECTION 2.2 THE LIMIT OF A FUNCTION
ANSWERS 1. The answer, of course, depends on the starting digit: x 01 0 01 0 001 0 0001 0 00001 0 000001
sin
π x
0 0 0 0 0 0
π x
sin
x 02 0 02 0 002 0 0002 0 00002 0 000002
sin
x
0 0 0 0 0 0
03
3 2
0 03
3 2 3 2
0 003
3 2 3 2 3 2
0 0003 0 00003 0 000003
x 04 0 04 0 004 0 0004 0 00004 0 000004
x 07 0 07 0 007 0 0007 0 00007 0 000007
π sin x 1 0 0 0 0 0
x 05 0 05 0 005 0 0005 0 00005 0 000005
π x 0 974927912 0 781831482 0 433883739 0 974927912 0 781831482 0 433883739 sin
π sin x 0 0 0 0 0 0
π x
sin
x
2 2
08 0 08 0 008 0 0008 0 00008 0 000008
1 0 0 0 0
y 1
1 x
_1 59
3 2
0 06
3 2
0 006
3 2
0 0006
3 2
0 00006
3 2
0 000006
3 2
09 0 09 0 009 0 0009 0 00009 0 000009
4. There is no limit.
π x
06
x
3. Answers will vary.
0
sin
x
2. Answers will vary.
π x
π x 0 342020143 0 342020143 0 342020143 0 342020143 0 342020143 0 342020143 sin
CHAPTER 2 LIMITS AND DERIVATIVES
HOMEWORK PROBLEMS CORE EXERCISES 1, 5, 8, 11, 19, 33, 50 SAMPLE ASSIGNMENT 1, 5, 7, 8, 11, 16, 17, 19, 33, 42, 50, 53 EXERCISE 1 5 7 8 11 16 17 19 33 42 50 53
D
A
60
N
G
GROUP WORK 1, SECTION 2.2 An Interesting Function 1. Create a graph of the function y
1
2. Estimate lim
1 , 1 21 x
2
x
2.
from the graph. Back up your estimate by looking at the function, and discussing
1 21 x why your estimate is probably correct. x
0
1
3. Estimate lim
from the graph. Back up your estimate by looking at the function, and discussing
1 21 x why your estimate is probably correct. x
0
1
4. Does lim x
0
1
21
exist? Justify your answer. x
61
GROUP WORK 2, SECTION 2.2 Infinite Limits 1. Draw an odd function which has the lines x
and x 2
2. Analyze the vertical asymptotes of
3x 2
4x
16x 4
5 81
3 2
.
62
among its vertical asymptotes.
GROUP WORK 3, SECTION 2.2 The Shape of Things to Come In this activity we are going to explore a set of functions: f1 x
x f2
x
x 2 f3
x
fn x
x3
x n , n any positive integer
1. To start with, let’s practice the new notation. Compute the following: f1 2
f2 0 6
f3 0 8
f4 0
f 8 1 01
2. Sketch the functions f1 , f 2 , f3 , f 6 , and f8 on the set of axes below.
3. The number 0 plays a special role, since f n 0 0n a 0 such that f n a a for all positive integers n.
4. We know that lim f n x x
0
0 for all positive integers n. Find another number
0 for all positive integers n. How is this fact reflected on your graphs above?
63
GROUP WORK 3, SECTION 2.2 The Shape of Things to Come: Approaching Infinity 1. Using what you know about limits, compute the following quantities: (a) lim f 3 x x
(b) lim f 4 x
0
x
1
(c) lim f 15 x x
1
2. Using what you know about limits, compute the following quantities: 1
(a) lim f n
(b) lim f n 0 99
2
n
(c) lim f n x , where x n
3. Let g x
lim f n x for 0
n
n
1
(d) lim f n 1 n
x
1. Sketch g x , paying particular attention to g 1 and values of x
close to 1.
4. Are the following quantities defined? If so, what are they? If not, why not? lim f n x
(a) lim n
x
(b) lim
1
x
64
1
lim f n x n
GROUP WORK 3, SECTION 2.2 The Shape of Things to Come: The Nitty-Gritty By definition, “ lim f 2 x x
0” means that by taking x very close to zero, we can make x 2
0 smaller than
0
any small number you can name. 1. Find a number δ
0 such that if
2. Use a graph to find a number δ
δ
x
0 such that if
3. Now use algebra to find a number δ
1 100.
δ, then f 2 x
δ
0 such that if
x
1
δ, then x 2
δ
x
1
1
δ, then x 2
1 100.
1
1 100.
4. When constructing this problem,
1 100
was used as an arbitrary smallish number. Could you have done the
previous problems if we replaced
1 100
by
1 10 ,000 ?
How about
1 ? 1 ,000,000
5. Reread the first sentence on this page. How do your answers to Problems 1 and 4 show that lim f 2 x x
0? Do your answers to Problems 2, 3, and 4 show that lim x 2 x
65
1
1
0? Why?
0
GROUP WORK 4, SECTION 2.2 Why Can’t We Just Trust the Table? We are going to investigate lim sin x
0
π . We will take values of x closer and closer to zero, and see what value x
the function approaches. 1. Your teacher has given you a digit — let’s call it d . Fill out the following table. If, for example, your digit is 3, then you would compute sin , sin , sin , sin , etc. 03
0 03
0 003
0 0003
x
sin
π x
sin
π x
0d 0 0d 0 00d 0 000d 0 0000d 0 00000d 2. What is lim sin x
0
π ? x
3. Now fill out the table with a different digit. x 0d 0 0d 0 00d 0 000d 0 0000d 0 00000d Do you get the same result?
4. What is lim sin x
0
π ? x 66
2.3
Calculating Limits Using the Limit Laws
SUGGESTED TIME AND EMPHASIS 1 class
Essential material
POINTS TO STRESS 1. The algebraic computation of limits: manipulating algebraically, examining left- and right-hand limits, using the limit laws to break monstrous functions into pieces, and analyzing the pieces. 2. The evaluation of limits from graphical representations. 3. Examples where limits don’t exist (using algebraic and graphical approaches). 4. The computation of limits when the limit laws do not apply, and the use of direct substitution property when they do. QUIZ QUESTIONS π?
TEXT QUESTION In Example 4, why isn’t lim g x x
1
ANSWER Because the limit isn’t affected by the function when x x 2 2ax a 2 DRILL QUESTION If a 0, find lim . x a x 2 a2 (A)
1 2a
(B )
1 2a 2
1 2a 2
( C)
1 only when x is near 1.
( D) 0
(E) Does not exist
ANSWER (D) MATERIALS FOR LECTURE Discuss why lim [[x ]] sin x x
0 is not a straightforward application of the Product Law.
0
x and determine why we cannot compute lim
Have the students determine the existence of lim x
x
0
Use the Squeeze Theorem to show that lim x 2 [[x ]] x
0
x.
0.
0
WORKSHOP/DISCUSSION x2 2 x
4 x3 , lim 2 x 0 x
Compute some limits of quotients, such as lim x
8 x3 , lim 2 x 3 x
8 x3 , and lim x 2 x 2
8 , always 2
attempting to plug values in first. Have the students check if lim x
x
lim
5
x x
5
x
5
x
5
exists, and then compute left- and right-hand limits. Then check
52 . 5 x
Do some subtle product and quotients, such as lim x
0
x
x sin x and lim x
1
3 x
1
2 .
Present some graphical examples, such as lim f x and lim f x in the graph below. x
0
x
y 67
2
1
y=f(x) 2
x
68
CHAPTER 2 LIMITS AND DERIVATIVES
GROUP WORK 1: EXPLORING LIMITS Have the students work on this activity in groups. Problem 2 is more conceptual than Problem 1, but makes an important point about the sums and products of limits. ANSWERS 1. (a) (i) Does not exist (b) (i) Does not exist
(ii) Does not exist (ii) 1
(c) (i) 0
(iii) 4
(iv) Does not exist
(ii) Does not exist
2. (a) Both quantities exist. (b) Each quantity may or may not exist. GROUP WORK 2: FIXING A HOLE This activity foreshadows concepts used later in the discussion of continuity, in addition to giving the students practice in taking limits. After the activity, point out that mathematicians use the word “puncture” as well as “hole”. ANSWERS 1. No, yes, yes, no 2. x x
2
3. Does not exist, 13 4.
y
1 1
x
One of the discontinuities can be “filled in” and the other cannot. 5. A “hole” is an x -value at which the function is not defined, yet the left- and right-hand limits exist. Or: A “hole” is an x -value where the function is undefined, yet the function is defined near x . Or: A “hole” is an x -value at which we can add a point to the function and thus make it continuous there. 6.
y 1
1
g has a hole at x
0. 68
x
SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS
GROUP WORK 3: THE SQUEEZE THEOREM sin x 0 x
This activity gives an informal graphical way to show that lim x
1. A more careful geometric argument
is given in Section 3.3. ANSWERS 1, 2.
y y=x
1
3. For x
0, sin x
x
For x
0, sin x
x
y=sin x y=x-x#/6
x sin x x
1. 1
(reversing the second inequality because x 0). 0
_1
x 6 5. The Squeeze Theorem now gives lim 1. x 0 x
1 x
4. f x
_1
HOMEWORK PROBLEMS CORE EXERCISES 2, 5, 18, 50, 51, 60 SAMPLE ASSIGNMENT 2, 5, 10, 18, 32, 35, 47, 50, 51, 60, 61
EXERCISE 2 5 10 18 32 35 47 50 51 60 61
D
A
69
N
G
1
GROUP WORK 1, SECTION 2.3 Exploring Limits 1. Given the functions f and g (defined visually below) and h and j (defined algebraically), compute each of the following limits, or state why they don’t exist: y
y f
x x
h x
g
1
1
1 0
2
(a) (i) lim f x x
2
x
2
x
h x
2
x
(iv) lim
2
2
2
j x
(ii) lim f x j x
2
x
2
2. (a) In general, if lim m x exists and lim n x exists, is it true that lim [m x x
2
j x x
(ii) lim f x
(c) (i) lim f x g x x
(iii) lim h x
if x
x
2
0
(ii) lim g x
(b) (i) lim g x x
x
4 2
a
x
a
x
a
n x ] exists? How
about lim [m x n x ]? Justify your answers. x
a
(b) In general, if lim m x does not exist and lim n x does not exist, is it true that lim [m x x
a
x
a
x
a
does not exist? How about lim [m x n x ]? Compare these with your answers to part (a). x
a
70
n x ]
GROUP WORK 2, SECTION 2.3 Fixing a Hole x
Consider f x
2
x2
x
2
1. Is f x defined for x
. 1? For x
0? For x
1? For x
2?
2. What is the domain of f ?
x
3. Compute lim x2
x
4. Graph y
x x2
2 x
2 x
2
x
and lim 2
x
2
x2
2 x
. Notice that one limit exists, and one does not. 2
. There are two x -values that are not in the domain of f . Later, we will call these
“discontinuities”. Geometrically, what is the difference between the two discontinuities?
5. We say that f x has one hole in it. Where do you think that the hole is? Define “hole” in this context. sin x 6. The function g x is not defined at x 0. Sketch this function. Does it have a hole at x 0? x
71
GROUP WORK 3, SECTION 2.3 The Squeeze Theorem sin x . 0 x
In this activity, we take a graphical approach to computing lim x
1. Using a graphing calculator, show that if 0
x
1, then x
x3 6
sin x
x . Give a rough sketch of the
three functions over the interval [0 1] on the graph below.
y 1
1
x
x3 sin x 6 done so already, add these portions of the three functions to your graph above.
2. Again using a graphing calculator, show that if
3. Explain why
sin x x
1 for
1
x
1, x
1
x
0, then x
0. Use the inequalities in parts 1 and 2 to help you.
4. Again using parts 1 and 2, can you find a function f x with f x that lim f x x
0
1?
5. Using parts 3 and 4, compute lim x
0
x . If you have not
sin x . x
72
sin x on x
1
x
1, x
0 such
The Precise Definition of a Limit
2.4
SUGGESTED TIME AND EMPHASIS 1–1 21 classes
Optional material
POINTS TO STRESS 1. The geometry of the ε-δ definition, what the notation means, and how it relates to the geometry. 2. The “narrow range” definition of a limit, as defined below. 3. Extending the precise definition to one-sided and infinite limits.
QUIZ QUESTIONS TEXT QUESTION Example 1 finds a number δ such that x 3 Why does this not prove that lim x 3 5x 6 2? x
5x
6
2
0 2 whenever x
δ.
1
1
ANSWER It is not a proof because we only dealt with ε
0 2; a proof would hold for all ε.
DRILL QUESTION Let f x 5x 2. Find δ such that f x ANSWER δ 0 002 works, as does any smaller δ.
12
0 01 whenever
δ
x
2
δ.
MATERIALS FOR LECTURE The “narrow range” definition of limit may be covered as a way of introducing the ε–δ definition to the students in a familiar numerical context. We say that lim f x L if for any y-range centered at L there x
a
is an x -range centered at a such that the graph is “trapped” in the window — that is, does not go off the top or the bottom of the window. The transition to the traditional definition can now be made easier by observing that the width of the y-range is 2ε and the width of the x -range is 2δ. If the students are familiar with graphing calculators, this definition can be illustrated with setting different viewing windows for a particular graph. Make sure the students understand that limit proofs, as described in the book, are two-step processes. The act of finding δ is separate from writing the proof that the students’ choice of δ works in the limit definition. This fact is stated clearly in the text, but it is a novel enough idea that it should be reinforced. Discuss how close x needs to be to 4, first to ensure that
1 x
1
20 000. Then argue intuitively that lim 4
2
x
4
x
x
4
2
1000, and then so that
. 4
Using the formal definition of limit, show that neither 1 nor 73
1
2
1 is the limit of h x
1 1
if x if x
0 0
as x goes to 0. Emphasize that although this result is obvious from the graph, the idea is to see how the definition works using a function that is easy to work with.
75
CHAPTER 2 LIMITS AND DERIVATIVES
WORKSHOP/DISCUSSION Estimate how close x must be to 0 to ensure that sin x x is within 0 03 of 1. Then estimate how close x must be to 0 to ensure that sin x x is within 0 001 of 1. Describe what you did in terms of the definition of a limit. Return to the interesting function f x the right- and left-hand limits exist at x
1
1 21
x
from Group Work 1 in Section 2.2, and describe why
0, but the limit does not exist.
Discuss why f x [[x ]] does not have a limit at x 0, first using the “narrow range” definition of limit, and then possibly the ε-δ definition of limit. ex 1 Then discuss why lim 1, using the x 0 x
y 2
1
“narrow range” definition of limit and a graph like the one at right. _1
0
1 x
Find, numerically or algebraically, a δ 0 such that if 0 x 0 δ, then x 3 compute a δ 0 such that if 0 x 2 δ, then x 3 8 10 3 .
0
3
10
. Similarly,
GROUP WORK 1: A JITTERY FUNCTION This activity can be done in several ways. After they have worked for a while, perhaps ask one group to try to solve it using the Squeeze Theorem, another to solve it using the “narrow range” definition of limit, and a third to solve it using the ε-δ definition of limit. They should show why their method works for Problem 2, and fails for Problem 3. ANSWERS 1.
2. lim f x
y
0. Choose ε with ε
0. Let δ
ε. Now
1
if δ x δ, then x 2 ε, regardless of whether x is rational or irrational. This can also be shown using the Squeeze Theorem and the fact that 0
f x
x 2 , and
then using the Limit Laws to compute lim 0 and lim x 2 . x
3. It does not exist. Assume that lim f x x
_1
0
1
x
10 1
ε
L must than
1 10 .
x
0
L . Choose
. Now, whatever your choice of δ, there are some
x -values in the interval 1 1. be less than 10
1
0
But there are also values of x in the interval with f x
δ 1
δ with f x 2 10, so
0, so
L must be greater
So L cannot exist. The “narrow range” definition of limit can also be used to solve this problem. 74
4. We can conjecture that the limit does not exist by applying the reasoning from Problem 3.
76
SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT
GROUP WORK 2: THE DIRE WOLF COLLECTS HIS DUE The students will not be able to do Problem 1 with any kind of accuracy. Let them discover for themselves how deceptively difficult it is, and then tell them that they should do the best that they can to show what is happening as x goes to zero. Ask them to compare their result with lim x sin π x . If a group finishes early, x
0
pass out the supplementary problems. ANSWERS 1.
y 1
_1
1
3 x
2
_1
2. (a) 1, 1, 1
(b) 1
(c) 0
(d) A function must approach only one number for the limit to exist.
ANSWERS TO SUPPLEMENTARY PROBLEMS 1. The length of the boundary is infinite. There are infinitely many wiggles, each adding at least 2 to the total perimeter length. 2. The area is finite. It is less than the area of the rectangle defined by 0
x
1,
2
y
1.
3. Answers will vary. GROUP WORK 3: INFINITY IS VERY BIG The precise definition of infinite limits is similar to the standard definition, but it is different enough that most students need a little practice before they can grasp it. ANSWERS 1. x
0 001
2. (a) Choose M . Now let δ
1 M
. If 0
x
1 M
, then
1 x2
M . Values of δ less than
1 M
work,
too. 1 (b) is large negative for small negative values of x , and large positive for small positive values of x . x GROUP WORK 4: THE SIGNIFICANCE OF THE ”FOR EVERY” The purpose of this activity is to allow the students to discover that rigor in mathematics is often necessary and useful. Problem 1 is designed to lead the students to make a false assumption about the third function, h x . Problem 2 should dispel that assumption. 75
CHAPTER 2 LIMITS AND DERIVATIVES
This activity is longer than it appears. Allow the students plenty of time to do the first three questions, which should help them to internalize and understand the formal definition of a limit. Closure is important to ensure that the “punchline” isn’t lost in the algebra. When the students are finishing up, it is crucial to pass out Problem 2. This part asks them to look at the functions a third time, with ε 0 01 Make sure that the students remember to check values of h x for x 0 and for x 0. Finish up by having them draw a graph of h x NOTE If time is limited, allow the students to find a δ that works from looking at graphs, as opposed to finding the largest possible δ algebraically. ANSWERS PART 1 1 4
1. (a) δ 1
2. δ
20
(b) δ 1
,δ
10
1 2
(c) Any δ will work. 0 08 if x
1
, any δ will work. h x
25
0
0
if x
0
which is always less than
1
10
.
3. Students may or may not see the wrinkle in h x at this point. PART 2 δ
1 200
,δ
1
, no δ will work. h x
100
1
0 08 if x
0
25
0
if x
0
N
G
HOMEWORK PROBLEMS CORE EXERCISES 3, 7, 28, 42 SAMPLE ASSIGNMENT 3, 7, 28, 33, 37, 41, 42, 44 EXERCISE 3 7 28 33 37 41 42 44
D
A
76
which is always greater than 0 01.
GROUP WORK 1, SECTION 2.4 A Jittery Function Not all functions that occur in mathematics are simple combinations of the “toolkit” functions usually seen in calculus. Consider this function: f x
0 x2
if x is rational if x is irrational
1. It is obvious that you can’t graph this function in the same literal way that you would graph y cos x , but it is useful to have some idea of what this function looks like. Try to sketch the graph of y f x .
2. Does lim f x exist? If so, what is its value? If not, why not? Make sure to justify your answer x
0
carefully.
3. Does lim f x exist? Carefully justify your answer. x 1
4. What do you conjecture about lim f x if a x
a
0?
77
GROUP WORK 2, SECTION 2.4 The Dire Wolf Collects his Due In this activity we will explore a function that is particularly loved by mathematicians everywhere, sin π x . 1. Sketch the graph of y
sin π x on the interval
2. It appears that this function is not defined at x even have a right-hand limit. (a) Evaluate sin π x at x
2 2 1 , 5,
(b) Evaluate sin π x for x
4n
(c) Evaluate sin π x for x integer.
2
3].
0 does not have a limit at x
0 and in fact, does not
and 29 .
1,
1 1 1 , 2,
n a positive integer, using the pattern from part (a).
and 13 . Using this pattern, evaluate sin π x for x
(d) Give an argument to show that lim sin π x does not exist. x
0
78
1 n,
n a positive
GROUP WORK 2, SECTION 2.4 The Dire Wolf Collects his Due (Supplementary Problems) Consider the region bounded on the bottom by the line y on the left by the line x the line x 1, and on top by the graph of y sin π x as shown: y 1
1 x
0
_1
_2
1. Is the length of the boundary of this region finite or infinite? Justify your answer.
2. Is the area of this region finite or infinite? Justify your answer.
3. Do you think this result is as interesting as we do? Why or why not?
79
0, on the right by
GROUP WORK 3, SECTION 2.4 Infinity is Very Big 1. For what values of x near 0 is it true that
2. The precise definition of lim f x x
a
1 x2
1,000,000?
states that for every positive number M , no matter how large,
there is a corresponding positive number δ such that f x 1 (a) Use this definition to prove that lim x 0 x2
(b) Why is it not true that lim x
0
1 x
M whenever 0
? Give reasons for your answer.
80
x
a
δ.
GROUP WORK 4, SECTION 2.4 The Significance of the ‘‘ For Every’’ (Part 1) Consider the following functions: f x
2x
3
x2 x
g x
4 2
x 25x
h x
We want to try to prove the following statements: lim f x x
1
5
lim g x x
4
lim h x x
2
1 25
0
Notice that these are not obvious statements, since g 2 and h 0 are both undefined. 1. We start with ε
1 2.
(a) Can you find a number δ with the property that, when x answer with a graph, and prove it algebraically.
1
δ, f x
(b) Can you find a number δ with the property that, when x
2
δ, g x
(c) Can you find a number δ with the property that, when x
0
δ, h x
81
5
4
1 25
1 2?
1 2?
1 2?
Illustrate your
The Significance of the ‘‘ For Every’’ (Part 1)
2. We now have some reason to believe that the above statements are true. But just having “some reason to 1 believe” isn’t enough for mathematicians. Repeat the previous problem for ε 10.
3. Now, what do you believe about these limits?
82
GROUP WORK 4, SECTION 2.4 The Significance of the ‘‘ For Every’’ (Part 2) Try the three limits again, this time for ε x x0 Part 1?
δ, you check values of x0
1 100 .
x and x0
Make sure that when you are trying to verify the condition x . Do you wish to change your answer to Problem 3 from
83
2.5
Continuity
SUGGESTED TIME AND EMPHASIS 1–1 12 classes
Essential material
POINTS TO STRESS 1. The graphical and mathematical definitions of continuity, and the basic principles. 2. Examples of discontinuity. 3. The Intermediate Value Theorem: mathematical statement, graphical examples, and applied examples. QUIZ QUESTIONS TEXT QUESTION The text says that y tan x is discontinuous at x 2 . This would seem to contradict Theorem 7. Does it? Why or why not? ANSWER It does not; tan x is indeed continuous at every point in its domain, but x 2 is not in its domain. DRILL QUESTION Assume that f 1 , and f 3 5. Does there have to be a value of x , between 1 and 3, such that f x 0? ANSWER No, there does not. Only if the function is continuous does the IVT indicate that there must be such a value. MATERIALS FOR LECTURE Discuss the idea of continuity at a point, continuity on an interval, and the basic types of discontinuities. Note that the statement “ f is continuous at x
a” is implicitly saying three things:
1. f a exists. 2. lim f x exists. x
a
3. The two quantities are equal. To show that all three statements are important to continuity, have the students come up with examples where the first holds and the second does not, the second holds and the first does not, and where the first two hold and the third does not. Examples are sketched below. y
y
y
x
x
x
Some students tend to believe that all piecewise functions are discontinuous at the border points. Examine x2 if x 1 the function f x 1 and x e. This would be a good ln x 1 if 1 x e at the points x x if x e time to point out that the function x is continuous everywhere, including at x 0 Start by stating the basic idea of the Intermediate Value Theorem (IVT) in broad terms. (Given a function on an interval, the function hits every y-value between the starting and ending y-values.) Then attempt 84
SECTION 2.5 CONTINUITY
to translate this statement into precise mathematical notation. Show that this process reveals some flaws in our original statement that have to be corrected (the interval must be closed; the function must be continuous.) To many students the IVT says something trivial to the point of uselessness. It is important to show examples where the IVT is used to do non-trivial things. Example: A graphing calculator uses the IVT when it graphs a function. A pixel represents a starting and ending y-value, and it is assumed that all the intermediate values are there. This is why graphing calculators are notoriously bad at graphing discontinuous functions. Example: Assume a circular wire is heated. Use the IVT to show that there exist two diametrically opposite points with the same temperature. ANSWER Let f x be the difference between the temperature at a point x and the temperature at the point opposite x . f is a difference of continous functions, and is thus continuous itself. If f x 0, then f x 0, so by the IVT there must exist a point at which f 0. Example: Show that there exists a number whose cube is one more than the number itself. (This is Exercise 69.) ANSWER Let f x x3 x there exists an x with f x
1. f is continuous, and f 0 0.
0 and f 2
0. So by the IVT,
0 x irrational p Have the students look at the function f x 1 x q , where p and q are integers, q is q positive, and the fraction is in lowest terms This function, discovered by Riemann, has the property that it is continuous where x is irrational, and not continuous where x is rational. WORKSHOP/DISCUSSION Indicate why f x csc x is continuous everywhere on its domain, but is not continuous everywhere. Then discuss the continuity of g x e csc x , and why all the discontinuities of g are removable. 0 if x is rational Ask If the group activity “A Jittery Function” was assigned, revisit f x x 2 if x is irrational the students to guess if this function is continuous at x 0. Many will not believe that it is. Now look at it using the definition of continuity. They should agree that f 0 0. In the activity it was shown that lim f x existed and was equal to 0. So, this function is continuous at x 0. A sketch such as the one x
0
found in the answer to that group work may be helpful. Present the following scenario: two ice fishermen are fishing in the middle of a lake. One of them gets up at 6:00 P. M . and wanders back to camp along a scenic route, taking two and a half hours to get there. The second one leaves at 7:00 P. M ., and walks to camp along a direct route, taking one hour to get there. Show that there was a time where they were equidistant from camp. Revisit Exercise 5 in Section 2.2, discussing why the function is discontinuous. 1 Show that f x is not continuous. (This is the same function used in “An Interesting 1 21 x 85
Function”, Group Work 1 in Section 2.2.)
87
CHAPTER 2 LIMITS AND DERIVATIVES
GROUP WORK 1: EXPLORING CONTINUITY Warm the students up by having them graph 2 x without their calculators, and asking where it is continuous. The first problem is appropriate for all classes. Problem 2 assumes the students have previously seen the activity “A Jittery Function”. If they have not, skip it and go directly to Problem 3. Before handing out Problem 3, make sure that the students recall the definition of the greatest integer (“floor”) function y [[x ]]. After this activity, discuss the continuity of [[x ]] at integer and at non-integer values. Problem 4 is intended for classes with a more theoretical bent. ANSWERS 1. c
4, m
5
2. (b) 0
(c) 0
3. (a)
(d) It is continuous because f 0
lim f x .
x
(b) All values except a
y
x2
(c) lim x
0
0; lim x
0
1, x2
2,
3, 2
does not exist because
2
the left- and right-hand limits are different. 1 0
x
1
4. (a) The fact that f is continuous implies that lim f x x
lim h x x
a
f x
lim x
2
a
(b) False. For example, let f x
2
lim f x x
a
f a
2
f a for all a. Then, by the Limit Laws, h a .
a
1
1 if x if x
0 0
GROUP WORK 2: THE AREA FUNCTION This activity is designed to reinforce the notion of continuity by presenting it in an unfamiliar context. It will also ease the transition to area functions in Chapter 5. It is important that this activity be well set up. Do Problem 1 with the students, making sure to compute a few values of A r and to sketch it. The students should try to answer Problems 2 and 3 using their intuition and the definition of continuity. It may be desirable to have the students restrict themselves to r 0. Note that in this activity, one can “prove” continuity by looking at the actual formulas for A r and B r , but that the goal of the activity is that the students understand intuitively why both area functions are continuous. Students may disagree on the answer to Problem 3. If you are fortunate enough to have groups that have reached opposite conclusions, break up one or more of them, and have representatives go to other groups to try to convince them of the error of their ways. ANSWER Yes to all three questions. For all r , A r , B r , and C r exist; and lim A x x
86
r
A r ,
lim B x
x
r
B r , and lim C x x
r
C r . (The limits can be shown to exist by looking at the left- and
right-hand limits.)
88
SECTION 2.5 CONTINUITY
GROUP WORK 3: THE TWIN PROBLEM When students see this problem, there is a good chance that they will disagree among themselves about the answer. Let them argue for a while. Ideally, they will come up with the idea of using the Intermediate Value Theorem to prove that Dr. Stewart was correct. If they don’t, this may need to be given to them as a hint. Another hint they may need is that the Intermediate Value Theorem deals with a single continuous function, whereas the problem is talking about two functions, Stewart’s temperature and Shasta’s temperature. They will have to figure out a way to find a single function that they can use. Encourage them to write up a solution to the exact degree of rigor that will be expected of them on homework and exams; this is a good opportunity to convey the course’s expectations to the students. ANSWER Let S t and O t be Dr. Stewart’s and Shasta’s temperatures at time t . Now let T t S t O t . T t is continuous (being a difference of continuous functions), T 0 0 (Dr. Stewart is warmer at first), and T f 0 (where f represents the end of the vacation; Shasta is warmer at the end). Therefore, by the IVT, there exists a time a at which T a 0 and hence S a O a . Notice that most students who try to argue that the conclusion is false (using things such as stasis chambers and exceeding the speed of light) are really trying to construct a scenario where the continuity of the temperature function is violated. GROUP WORK 4: SWIMMING TO THE SHORE Emphasize to the students that they are not trying to find x , but simply trying to prove its existence. As in the Twin Problem, a first hint might be to use the IVT, and a second could be to find a single continuous function of x . It is probably best to do this activity after the students have seen the solution to the ice fisherman problem above, or the Twin Problem. ANSWER Let D x d Px A d Px B . D x is continuous, D A IVT, there is a place where D x 0. HOMEWORK PROBLEMS CORE EXERCISES 4, 7, 10, 12, 24, 44, 53, 67 SAMPLE ASSIGNMENT 4, 7, 10, 12, 15, 19, 24, 25, 40, 44, 53, 67, 73 EXERCISE 4 7 10 12 15 19 24 25 40 44 53 67 73
D
A
87
N
G
0, and D B
0. Therefore, by the
GROUP WORK 1, SECTION 2.5 Exploring Continuity 1. Are there values of c and m that make h x
cx 2
if x
1
4
if x if x
1 1
x3
mx
continuous at x
1? Find c
and m, or explain why they do not exist.
2. Recall the function f x
0 x2
if x is rational if x is irrational
(a) Do you believe that f x is continuous at x
0? Why or why not?
(b) What is f 0 ?
(c) What is lim f x ? x
0
(d) Use parts (b) and (c) either to revise your answer to part (a), or to prove that your answer is correct.
88
Exploring Continuity
x2
3. Consider the function h x
(a) Sketch the graph of the function for
1
x
2.
y
1
0
(b) For what values of a,
1
a
1
2, is lim h x x
a
x
h a ?
(c) Compute lim x 2 and lim x 2 , if they exist. Explain your answers. x
0
x
2
89
Exploring Continuity
4. We know that the function g x
x 2 is continuous everywhere.
(a) Show that if f is continuous everywhere, then h x argument.
f x
2
is continuous everywhere, using a limit
(b) Is it true or false that if h x f x 2 is continuous everywhere, then f is continuous everywhere? If it is true, prove it. If it is false, give a counterexample.
90
GROUP WORK 2, SECTION 2.5 The Area Function The following are graphs of y
f x ,y
g x , and y
h x :
y
y
y g
f 2
h
1 0
1 0
x
f x
_1 1
1
2
g x
x
1 0
x
1
1
h x
2x x
2
if x 4 if x
x
1 1
1. Let A r be the area enclosed by the x -axis, the y-axis, the graph of the function f , and the line x Would you conjecture that A r is continuous at every point in the domain of f ? Why or why not?
r.
2. Let B r be the area enclosed by the x -axis, the y-axis, the graph of the function g, and the line x Would you conjecture that B r is continuous at every point in the domain of g? Why or why not?
r.
3. Let C r be the area enclosed by the x -axis, the y-axis, the graph of the function h, and the line x Would you conjecture that C r is continuous at every point in the domain of h? Why or why not?
r.
91
GROUP WORK 3, SECTION 2.5 The Twin Problem There is a bit of trivia about the author of your textbook, Dr. James Stewart, that few people know. He has an evil twin sister named Shasta. Although he loves his sister dearly, she dislikes him and tries to be different from him in all things. Last winter, they both went on vacation. Dr. Stewart went to Hawaii. Shasta had planned on going to Aruba, but she decided against it. She hates her brother so much that she was afraid there would be a chance that they might be experiencing the same temperature at the same time, and that prospect was distasteful to her. So she decided to vacation in northern Alaska. After a few days, Dr. Stewart received a call: “This is Shasta. I am cold and uncomfortable here. That’s good, since you are undoubtedly warm and comfortable, and I want us to be different. But I’m not sure why I should be the one in northern Alaska. I think we should switch places for the last half of our trip.” “It is only fair,” he agreed. So they each traveled again. Dr. Stewart took a trip from Hawaii to Alaska, while Shasta took a trip from Alaska to Hawaii. They each traveled their own different routes, perhaps stopping at different places along the way. Eventually, they had reversed locations. Dr. Stewart was shivering in Alaska; Shasta was in Hawaii, warm and happy. She received a call from her brother. “Hi, Shasta. Guess what? At some time during our travels, we were experiencing exactly the same temperature at the same time. So HA!” Is Dr. Stewart right? Has Good triumphed over Evil? He would try to write out a proof of his statement, but his hands are too frozen to grasp his pen. Help him out. Either prove him right, or prove him wrong, using mathematics.
92
GROUP WORK 4, SECTION 2.5 Swimming to the Shore A swimmer crosses a river starting at point A and ending at point B, following the path shown below. Prove that for some value x , the swimmer’s distance d Px A from A is the same as the distance d Px B from B. y
B
d(Px, B) Px d(Px, A) Swimmer’s Path A
0
a
x
b
93
x
Limits at Infinity; Horizontal Asymptotes
2.6
SUGGESTED TIME AND EMPHASIS 1 class
Essential material(This material may also be covered after Section 4.2.)
POINTS TO STRESS 1. The geometric and limit definitions of horizontal asymptotes, particularly as they pertain to rational functions. 2. The computation of infinite limits. 3. The technique and the dangers of using calculators to check limits (both numerically and graphically).
QUIZ QUESTIONS TEXT QUESTION To evaluate the limit at infinity of a rational function, we first divide both the numerator and denominator by the highest power of x that occurs in the denominator. Why must we do such a thing? ANSWER By doing this division, we make the denominator approach a finite value as x . Now we can take the limit of the numerator, and easily divide it by the limit of the denominator. DRILL QUESTION Compute lim x
1 x3
x2 5x 2
2x 3 . 3x 5
ANSWER 2
MATERIALS FOR LECTURE Describe asymptotes verbally and then give graphical and limit definitions. Note that a function can cross its horizontal asymptote. Explain the difference between the definitions of lim f x L and x
lim f x
x
a
L , emphasizing that in one case we choose a small δ and in the other, a large value N .
Perhaps include a description of slant asymptotes. Ask students if a function can be bounded but not have a horizontal asymptote. Does sin x have a horizontal sin x sin x asymptote? What about ? How is different? 94 100 x
SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
Examine lim ln ln x on a graphing calculator, first by plugging in large numbers, then by examining the x
graph. Then show that this limit is, in fact, infinity. If teaching an advanced class, one might try to “prove” that this limit is the expected 5 429 using epsilons and deltas, and see how the attempt fails. (NOTE 5 429 is ln ln 1099 , which is what a student would come up with by plugging very large numbers into a calculator.) y
y
y
6
6
6
5
5
5
4
4
4
3
3
3
2
2
2
1
1
1
0
2000
4000
6000
8000 10,000
y
ln ln x
x
0
2 × 10' 4 × 10' 6 × 10' 8 × 10' 10 × 10'
y
x
0
2 × 10' ' 4 × 10' ' 6 × 10' ' 8 × 10' ' 10 × 10'
y
ln ln x
x
ln ln x
Discuss rates of growth. For large values of x , 3x 2x x3 x2 x ln x ln ln x , even though x they all approach infinity. (An advanced class can discuss the even larger x .) Point out that functions such as 0 85x and x 2 don’t go to infinity. Note that for values of x near zero, x x2 x 3 , although all approach zero. Point out that as x approaches 0, a x approaches 1 and loga x approaches .
WORKSHOP/DISCUSSION Compute the limits of y
x x2
5 as x 25
, and as x
. Graph the function. Also perhaps
review how to find the limits as x 2x 3 x3
Graph y
16 , after calculating limits as x 27
3 and as x
. x
x
Calculate lim e . Show the students how to find a domain for x such that e x
0 001 for all x in that
domain. Examine lim x
[[x ]] [[x ]] and lim 2 . x x x
GROUP WORK 1: TO INFINITY AND BEYOND This activity is intended to develop the students’ intuition about infinite limits. While they should justify their answers, it is important that they also get some feel for how limits as x behave. ANSWERS 1. (a) y
1 3
(b) None
(c) y
6
2. 0 95
3.
4. 0
CHAPTER 2 LIMITS AND DERIVATIVES
GROUP WORK 2: INFINITE LIMITS This activity is too long to be done in a 50-minute session. Pick and choose problems. It is more important to have good introduction and closure on each part than to have all of them worked out. Problem 4 is an extension of Exercise 55. ANSWERS 3
y
3
3. Answers will vary. Possible answers: (a) f x g x
x
g x 4. (a)
1
x
(b) f x
x, g x
(d) f x
x2
(b) 0
bn
(c)
x2
(c) f x
x 2, x,
x2
42, g x
or
x
2
GROUP WORK 3: I AM THE GREATEST Before handing this activity out, make sure the students know the definition of the greatest integer function, and can sketch its graph. ANSWERS 1. This can be done from the graph, or using the definition. (Choose ε
0, then let δ
2. (a) y 5 4 3 2 1 0
(b) Lower bound
n n
1
1 1 1 6 5 4
1 3
1 2
3. 0 1,
1 x
x
, upper bound 1
(c) Use the Squeeze Theorem, taking the limits of the bounds as n
4. When x
1
0 96
0.
0 2.)
SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
5. lim x
[[x ]] x
1. This can be seen by a similar bounding argument to the one above. If you use this activity,
it is a good idea to show the graph to your students, for it is a truly pretty thing: y
y
y
1
1
1
1
2
3
4
5
The tops of the lines are at y
0
x
2
4
6
8
10 x
1 and the bottoms trace out the curve y
HOMEWORK PROBLEMS CORE EXERCISES 3, 10, 49, 55, 56, 71, 77 SAMPLE ASSIGNMENT 3, 10, 12, 18, 44, 49, 51, 55, 56, 59, 65, 68, 71, 77, 81 EXERCISE 3 10 12 18 44 49 51 55 56 59 65 68 71 77 81
D
A
97
N
G
0
1
10
1 x.
20
30
40
50 x
GROUP WORK 1, SECTION 2.6 To Infinity and Beyond 1. Describe the horizontal asymptotes, if any, of the following functions. (a) f x
x4 3x 4
(b) f x
2x 5 2x 3 18 x4 x2 x 2
(c) f x
2x 5 2x 3 18 x 4 3x 3 x 2
2. Find lim x 25 e x
3. Find lim x
4. Find lim x
x2 2 x2 5
2x
x.
x . ln x
cos x
.
ln ln x
98
GROUP WORK 2, SECTION 2.6 Infinite Limits 1. Draw an even function which has the lines y
1, x
4, and x
2. Describe all vertical and horizontal asymptotes of f x
3x 2
1 among its asymptotes.
4x
16x 4
3. Find formulas for two functions, f and g, such that lim f x f x
g x
(b) lim
f x
g x
(c) lim
f x
g x
0
(d) lim
f x
g x
42
x
x
x
x
4. Let P x am x m n, respectively. (a) Find lim x
(b) Find lim x
(c) Find lim x
P x
a1 x
if m
n.
if m
n.
if m
n.
a0 , and Q x
bn x n
Q x
P x Q x
P x Q x
99
81
.
x
x
(a) lim
5
b1 x
lim g x and
b0 be polynomials of degree m and
GROUP WORK 3, SECTION 2.6 I Am the Greatest 1. Show that lim x
0
[[x ]] x
0
2. (a) Sketch a graph of
1 x
on the axes below. y 5 4 3 2 1 0
(b) If
1 n
1
x
0.2
0.4
0.6
0.8
1.0 x
1 find upper and lower bounds for the expression x n
(c) Use the estimates above to show that lim x x
0
1 x
100
1.
1 . x
I Am the Greatest
3. Compute lim x 2 x
0
4. Show that lim x x
5. Compute lim x
1
1 . x
1 x
0.
[[x ]] Justify your reasoning.
x
101
2.7
Derivatives and Rates of Change
SUGGESTED TIME AND EMPHASIS 1–2 classes
Essential material
POINTS TO STRESS 1. The slope of the tangent line as the limit of the slopes of secant lines (visually, numerically, algebraically). 2. Physical examples of instantaneous rates of change (velocity, reaction rate, marginal cost, and so on) and their units. f x f a f a h f a . and f a lim 3. The derivative notations f a lim x a x a h 0 h
4. Using f to write an equation of the tangent line to a curve at a given point. 5. Using f as an approximate rate of change when working with discrete data.
QUIZ QUESTIONS TEXT QUESTION Why is it necessary to take a limit when computing the slope of the tangent line? ANSWER There are several possible answers here. Examples include the following: By definition, the slope of the tangent line is the limit of the slopes of secant lines. You don’t know where to draw the tangent line unless you pick two points very close together. The idea is to get them thinking about this question. DRILL QUESTION For the function g whose graph is given, arrange the following numbers in increasing order and explain your reasoning: 0
g
g 0
g 2
y y=g(x)
_1
ANSWER g 0
0
g 4
g
2
0
1
g 2 102
2
3
4
x
g 4
SECTION 2.7 DERIVATIVES AND RATES OF CHANGE
MATERIALS FOR LECTURE Review the geometry of the tangent line, and the concept of “locally linear”. Estimate the slope of the line tangent to y x 3 x at 1 2 by looking at the slopes of the lines between x 0 9 and x 1 1, x 0 99 and x 1 01, and so forth. Illustrate these secant lines on a graph of the function, redrawing the figure when necessary to illustrate the “zooming in” process. y 2.2
y 3
y 2.4
2.1
2.2
2
2
2 1
0
1.8
1.9
1
2
Similarly examine y
3
1 x2 1
x
0
0.9
1
1.1
0
1.2 x
0.8
1
1.2
1.4
x
at 0 1 .
y
y 1.1
y
2 1.4 1.5
1.2 1
1
1
0.5
0.8 0.9
0.6 _1
_0.5
0
0.5
1 x
_0.4 _0.2
0
0.2
0.4
If “A Jittery Function” was covered in Section 2.4, look at f x
_0.1
0 x2
_0.05
0
0.05
0.1 x
if x is rational if x is irrational
Poll the
class: Is there a tangent line at x 0? Then examine what happens if you look at the limits of the secant lines. Have students estimate the slope of the tangent line to y sin x at various points. Foreshadow the concept of concavity by asking them some open-ended questions such as the following: What happens to the function when the slope of the tangent is increasing? Decreasing? Zero? Slowly changing? Discuss how physical situations can be translated into statements about derivatives. For example, the budget deficit can be viewed as the derivative of the national debt. Describe the units of derivatives in real world situations. The budget deficit, for example, is measured in billions of dollars per year. Another example: if s d represents the sales figures for a magazine given d dollars of advertising, where s is the number of magazines sold, then s d is in magazines per dollar spent. Describe enough examples to make the pattern evident. Note that the text shows that if f x x 2 8x 9, then f a 2a 8. Thus, f 55 102 and f 100 192. Demonstrate that these quantities cannot be easily estimated from a graph of the function. Foreshadow the treatment of a as a variable in Section 2.8. If a function models discrete data and the quantities involved are orders of magnitude larger than 1, we can use the approximation f x f x 1 f x . (That is, we can use h 1 in the limit definition of the derivative.) For example, let f t be the total population of the world, where t is measured in years since 1800. Then f 211 is the world population in 2011, f 212 is the total population in 2012, and f 103
211 is approximately the change in population from 2011 to 2012. In business, if f n is the total cost of producing n objects, f n approximates the cost of producing the n
105
1 th object.
CHAPTER 2 LIMITS AND DERIVATIVES
WORKSHOP/DISCUSSION “Thumbnail” derivative estimates: graph a function on the board and have the class call out rough values of the derivative. Is it larger than 1? About 1? Between 0 and 1? About 0? Between 1 and 0? About ? Smaller than 1? This is good preparation for Group Work 2 (“Oiling Up Your Calculators”). Draw a function like the following, and first estimate slopes of secant lines between x a and x b, and between x b and x c. Then order the five quantities f a , f b , f c , m P Q , and m Q R in decreasing order. [Answer: f b
mPQ
mQR
f c
f a .]
y R
P Q
a
b
c
x
Start the following problem with the students: A car is travelling down a highway away from its starting location with distance function d t
8 t3
6t 2
12t , where t is in hours, and d is in miles.
1. How far has the car travelled after 1, 2, and 3 hours? 2. What is the average velocity over the intervals [0 1], [1 2], and [2 3]? Consider a car’s velocity function described by the graph below. v
0
A
B
C
D
t
1. Ask the students to determine when the car was stopped. 2. Ask the students when the car was accelerating (that is, when the velocity was increasing). When was the car decelerating? 3. Ask the students to describe what is happening at times A, C , and D in terms of both velocity and acceleration. What is happening at time B? Estimate the slope of the tangent line to y
sin x at x
x
sin x
0 05 09 0 99 0 999 1 0001 1 001
0 0 4794 0 7833 0 8360 0 8409 0 8415 0 8420 104
1 by looking at the following table of values. sin x sin 1 x 1 0 841471 0 724091 0 581441 0 544501 0 540723 0 540260 0 539881
SECTION 2.7 DERIVATIVES AND RATES OF CHANGE
Demonstrate some sample computations similar to Example 4, such as finding the derivative of f t
1
t at t
3, or of g x
x
x 2 at x
1.
GROUP WORK 1: FOLLOW THAT CAR 8 t3
Start this problem by giving the students the function d t
6t 2
12t and having them sketch its
graph. Ask them how far the car has traveled after 1, 2, and 3 hours, and then show them how to compute the average velocity for [0 1], [1 2], and [2 3]. ANSWERS 1. y
2. It appears to stop at t
2.
3. 8 mi h, 2 mi h, 0 08 mi h
60
4. 0 mi h. This is where the car stops. 40 20 0
1
x
2
GROUP WORK 2: OILING UP YOUR CALCULATORS As long as the students have the ability to estimate the slope of a curve at a point, this is a good time to hint at the uniqueness of e as the base of an exponential function. ANSWERS 1. If the students do this numerically, they should be able to get some pretty good estimates of ln 3 1 098612. If they use graphs, they should be able to get 1 1 as an estimate. 2. 0 7 is a good estimate from a graph, and ln 2 0 693147 is attainable numerically. 3. As a increases, the slope of the curve at x 0 is increasing, as can be seen below.
_1
y
y
y
y
3
3
3
3
2
2
2
2
1
1
1
1
0
1
x
_1
0
1
x
_1
0
1
x
_1
0
1
x
4. The slope is less than 1 at a 2 and greater than 1 at a 3. Now apply the Intermediate Value Theorem. 5. The students are estimating e and should get 2 72 at a minimum level of accuracy. GROUP WORK 3: CONNECT THE DOTS Closure is particularly important on this activity. At this point in the course, many students will have the impression that all reasonable estimates are equally valid, so it is crucial that students discuss Problem 4. If there is student interest, this table can generate a rich discussion. Can A ever be negative? What would that mean in real terms? What would A
mean in real terms in this instance? 105
CHAPTER 2 LIMITS AND DERIVATIVES
ANSWERS 1. A 3500 0 06 % $ It is likely to be an overestimate, because the function lies below its tangent line near p 3500. 2. After spending $3500, consumer approval is increasing at the rate of about 0 06 % for every additional dollar spent. 3. Percent per dollar 4. A $3550 0 06 % $. This is a better estimate because the same figures now give a two-sided approximation of the limit of the difference quotient. GROUP WORK 4: DERIVATIVES AND INVERSES If inverse functions were covered, this activity is an excellent way for students to synthesize the two concepts, and to gain intuition and understanding about what the derivative means in a real-world context. ANSWERS 1. f 1 is the time at which a given number of centimeters of rain have fallen. The domain is from 0 cm to the maximum total rainfall. The range is from midnight to the end of the storm. 2. (a) At 5:00 A . M ., 2 cm of rain has fallen. (b) 5 cm of rain has fallen at 2:00 A . M . (c) At 5 A . M ., the rain is falling at the rate of 0 5 cm h. (d) After 5 cm of rain has fallen, time is passing at a rate of one half hour per centimeter of rainfall. HOMEWORK PROBLEMS CORE EXERCISES 3, 7, 13, 14, 18, 23, 29, 33, 59 SAMPLE ASSIGNMENT 3, 7, 11, 13, 14, 17, 18, 23, 29, 33, 37, 47, 49, 54, 59 EXERCISE 3 7 11 13 14 17 18 23 29 33 37 47 49 54 59
D
A
106
N
G
GROUP WORK 1, SECTION 2.7 Follow that Car Here, we continue with the analysis of the distance d t and t is in hours. 1. Draw a graph of d t from t
0 to t
8 t3
6t 2
12t of a car, where d is in miles
3.
2. Does the car ever stop?
3. What is the average velocity over [1 3]? over [1 5 2 5]? over [1 9 2 1]?
4. Estimate the instantaneous velocity at t
2. Give a physical interpretation of your answer.
107
GROUP WORK 2, SECTION 2.7 Oiling Up Your Calculators 1. Use your calculator to graph y method of your choosing.
3x . Estimate the slope of the line tangent to this curve at x
0 using a
2. Use your calculator to graph y method of your choosing.
2x . Estimate the slope of the line tangent to this curve at x
0 using a
3. It is a fact that, as a increases, the slope of the line tangent to y continuous way. Geometrically, why should this be the case?
a x at x
4. Prove that there is a special value of a for which the slope of the line tangent to y
0 also increases in a
a x at x
5. By trial and error, find an estimate of this special value of a, accurate to two decimal places.
108
0 is 1.
GROUP WORK 3, SECTION 2.7 Connect the Dots A company does a study on the effect of production value p of an advertisement on its consumer approval rating A. After interviewing eight focus groups, they come up with the following data: Production Value $1000 $2000 $3000 $3500 $3600 $3800 $4000 $5000
Consumer Approval 32% 33% 46% 55% 61% 65% 69% 70%
Assume that A p gives the consumer approval percentage as a function of p.
1. Estimate A $3500 . Is this likely to be an overestimate or an underestimate?
2. Interpret your answer to Problem 1 in real terms. What does your estimate of A $3500 tell you?
3. What are the units of A p ?
4. Estimate A $3550 . Is your estimate better or worse than your estimate of A $3500 ? Why?
109
GROUP WORK 4, SECTION 2.7 Derivatives and Inverses Let f t be the number of centimeters of rainfall that has fallen on my porch since midnight, where t is the time in hours. 1. Describe the inverse function f
1
in words. What are the domain and range of f
2. Interpret the following in practical terms. Include units in your answers. (a) f 5
(b) f
5
(c) f 5
(d)
2
f
1
2
05
5
05
110
1?
WRITING PROJECT
Early Methods for Finding Tangents
The history of calculus is a fascinating and too-often neglected subject. Most people who study history never see calculus, and vice versa. We recommend assigning this section as extra credit to any motivated class, and possibly as a required group project, especially for a class consisting of students who are not science or math majors. The students will need clear instructions detailing what their final result should look like. For example, recommend a page or two about Fermat’s or Barrow’s life and career, followed by two or three technical pages describing the alternate method of finding tangent lines as in the project’s directions, and completed by a final half page of meaningful conclusion.
111
2.8
The Derivative as a Function
SUGGESTED TIME AND EMPHASIS 2 classes
Essential material
POINTS TO STRESS 1. The concept of a differentiable function interpreted visually, algebraically, and descriptively. 2. Obtaining the derivative function f by first considering the derivative at a point x , and then treating x as a variable. 3. How a function can fail to be differentiable. 4. Sketching the derivative function given a graph of the original function. 5. Second and higher derivatives QUIZ QUESTIONS TEXT QUESTION The previous section discussed the derivative f
a for some function f . This section discusses
the derivative f x for some function f . What is the difference, and why is it significant enough to merit separate sections? ANSWER a is considered a constant, x is considered a variable. So f a is a number (the slope of the tangent line) and f x is a function. 3
DRILL QUESTION Consider the graph of f x Differentiable at x 0? Why?
x . Is this function defined at x
0? Continuous at x
0?
y 2 1 _10
_5
0
5
10 x
_1 _2
ANSWER It is defined and continuous, but not differentiable because it has a vertical tangent. MATERIALS FOR LECTURE Ask the class this question: “If you were in a car, blindfolded, ears plugged, all five senses neutralized, what quantities would you still be able to perceive?” (Answers: They could feel the second derivative of motion, acceleration. They could also feel the third derivative of motion, “jerk”.) Many students incorrectly add velocity to this list. Stress that acceleration is perceived as a force (hence F ma) and that “jerk” causes the uncomfortable sensation when the car stops suddenly. Review definitions of differentiability, continuity, and the existence of a limit. Sketch f from a graphical representation of f x sketch f
x2
4 , noting where f does not exist. Then
from the graph of f . Point out that differentiability implies continuity, and not vice versa. 112
SECTION 2.8 THE DERIVATIVE AS A FUNCTION
Examine graphs of f and f aligned vertically as shown. If
y
you wish to foreshadow f , add its graph below.
f
Discuss what it means for f to be positive, negative or zero. Then discuss what it means for f to be increasing, decreasing or constant.
x y f»
x y f»» x
If the group work “A Jittery Function” was covered in Section 2.4, then examine the differentiability of 0 if x is rational at x 0 and elsewhere, if you have not already done so. f x 2 x if x is irrational Show that if f x m, then f m 1 x
x4 0.
x2
x
1, then f
5
x
0. Conclude that if f x is a polynomial of degree
WORKSHOP/DISCUSSION Estimate derivatives from the graph of f x sin x . Do this at various points, and plot the results on the blackboard. See if the class can recognize the graph as a graph of the cosine curve. Given the graph of f below, have students determine where f has a horizontal tangent, where f is positive, where f is negative, where f is increasing (this may require some additional discussion), and where f is decreasing. Then have them sketch the graph of f . y
x
has more exercises of this type using a wide variety of functions. ANSWER There is a horizontal tangent near x 0. f is positive to the right of 0, negative to the left. f is increasing between the x -intercepts, and decreasing outside of them. TE C
y f»
x f
Compute f x and g x if f x x 2 x 2 and g x x 2 x 4. Point out that f x g x and 2 discuss why the constant term is not important. Next, compute h x if h x x 2x 2. Point out that 113
CHAPTER 2 LIMITS AND DERIVATIVES
the graph of h x is just the graph of f x shifted up one unit, so the linear term just shifts derivatives. TE C contains more explorations on how the coefficients in polynomials and other functions affect first and second derivatives. Consider the function f x x Show that it is not differentiable at 0 in two ways: by inspection , (it has a cusp); and by computing the left- and right-hand limits of f x at x 0 ( lim f x x
lim f x
x
0
).
0
TEC
TEC
can be used to develop students’ ability to look at the graph of a function and visualize the
graph of that function’s derivative. The key feature of this module is that it allows the students to mark various features of the derivative directly on the graph of the function (for example, where the derivative is positive or negative). Then, after using this information and sketching a graph of the derivative, they can view the actual graph of the derivative and check their work. GROUP WORK 1: TANGENT LINES AND THE DERIVATIVE FUNCTION This simple activity reinforces that although we are moving to thinking of the derivative as a function of x , it is still the slope of the line tangent to the graph of f . ANSWERS 1, 3. y
2. y
3
4. y
ln 2
1 x
2
2 ln 2 or y
ln 2
1 x
2
1 e
2 (2, g(2)) 1 y= g(x) 0
1
2
3
x
(1/e, g(1/e))
GROUP WORK 2: THE REVENGE OF ORVILLE REDENBACHER In an advanced class, or a class in which one group has finished far ahead of the others, ask the students to repeat the activity substituting “ D t , the density function” for V t . ANSWERS 1.
2.
y
3.
y
y
y = V (t ) y = V »»(t)
y = V »(t)
0 0
t
0
Units are cm3 s.
114
t
When the second derivative crosses the x -axis, the first derivative has a maximum, meaning the popcorn is
t
expanding the fastest.
116
SECTION 2.8 THE DERIVATIVE AS A FUNCTION
GROUP WORK 3: THE DERIVATIVE FUNCTION
Give each group of between three and five students the picture of all eight graphs. They are to sketch the derivative functions by first estimating the slopes at points, and plotting the values of f x . Each group should also be given a large copy of one of the graphs, perhaps on acetate. When they are ready, with this information they can draw the derivative graph on the same axes. For closure, project their solutions on the wall and point out salient features. Perhaps the students will notice that the derivatives turn out to be positive when their corresponding functions are increasing. Concavity can even be introduced at this time. Large copies of the answers are provided, in case the instructor wishes to overlay them on top of students’ answers for reinforcement. Note that the derivative of graph 6 (y e x ) is itself. Also note that the derivative of graph 1 (y cosh x ) is not a straight line. Leave at least 15 minutes for closure. The whole activity should take about 45–60 minutes, but it is really, truly worth the time.
If a group finishes early, have them discuss where f is increasing and where it is decreasing. Also show that where f is increasing, f is positive, and where f is decreasing, f is negative.
ANSWER (larger answer graphs are included after the group work)
Graph 1
Graph 2
Graph 3
Graph 4
Graph 5
Graph 6
Graph 7
Graph 8
115
CHAPTER 2 LIMITS AND DERIVATIVES
HOMEWORK PROBLEMS CORE EXERCISES 1, 3, 13, 16, 19, 28, 39, 42, 49 SAMPLE ASSIGNMENT 1, 3, 13, 14, 16, 17, 19, 22, 28, 36, 39, 42, 49, 61, 63 EXERCISE 1 3 13 14 16 17 19 22 28 36 39 42 49 61 63
D
A
116
N
G
GROUP WORK 1, SECTION 2.8 Tangent Lines and the Derivative Function The following is a graph of g x
x ln x . y 3
2 y=g(x) 1
0
1
2
It is a fact that the derivative of this function is g x
ln x
3
1.
1. Sketch the line tangent to g x at x 2 on the graph above. 2. Find an equation of the tangent line at x 2.
3. Now sketch the line tangent to g x at x
1 e
4. Find an equation of the tangent line at x
1 . e
0 368.
117
x
GROUP WORK 2, SECTION 2.8 The Revenge of Orville Redenbacher 1. Consider a single kernel of popcorn in a microwave oven. Let V t be the volume in cm3 of the kernel at time t seconds. Draw a graph of V t , including as much detail as you can, up to the time that the kernel is taken from the oven.
2. Now sketch a graph of the derivative function V t . What are the units of V t ?
3. Finally, sketch a graph of V
t . What does it mean when this graph crosses the x -axis?
118
GROUP WORK 3, SECTION 2.8 The Derivative Function The graphs of several functions f are shown below. For each function, estimate the slope of the graph of f at various points. From your estimates, sketch graphs of f .
Graph 1
Graph 2
Graph 3
Graph 4
Graph 5
Graph 6
Graph 7
Graph 8
119
The Derivative Function
Graph 1
120
The Derivative Function
Graph 2
121
The Derivative Function
Graph 3
122
The Derivative Function
Graph 4
123
The Derivative Function
Graph 5
124
The Derivative Function
Graph 6
125
The Derivative Function
Graph 7
126
The Derivative Function
Graph 8
127
The Derivative Function
Answer 1
128
The Derivative Function
Answer 2
129
The Derivative Function
Answer 3
130
The Derivative Function
Answer 4
131
The Derivative Function
Answer 5
132
The Derivative Function
Answer 6
133
The Derivative Function
Answer 7
134
The Derivative Function
Answer 8
135
2
SAMPLE EXAM
Problems marked with an asterisk (*) are particularly challenging and should be given careful consideration. 1. Consider the following graph of f .
4 3 2 1 4 1
2
5
3
(a) What is lim f t ? lim f t ? lim f t ? lim f t ? t
0
t
0
t
t
2
(b) For what values of x does lim f t exist? t
x
(c) Does f have any vertical asymptotes? If so, where?
(d) Does f have any horizontal asymptotes? If so, where?
(e) For what values of x is f discontinuous?
2. Find values for a and b that will make f continuous everywhere, if f x
3x ax x2
1 b
136
if x if 2 if 5
2 x x
5
t
CHAPTER 2 SAMPLE EXAM
3. Find the vertical and horizontal asymptotes for f x
x
4. Consider the function f x
x2
a
1
x
1
1
, where a is a positive number.
4 . 3x 4
(a) What is the domain of f ?
(b) Compute lim f x , if this limit exists. x
4
(c) Is f continuous at x 4? Explain your answer by either proving that f is continuous at x telling how to modify f to make it continuous.
4 or
5. Let f be a continuous function such that f 1 and f 1 1. Classify the following statements as (A) Always true (B) Never true, or (C) True in some cases, false in others. Justify your answers. (a) f 0
0
(b) For some x with
1
x
1, f x
0 137
CHAPTER 2 LIMITS AND DERIVATIVES
(c) For all x with
1
(d) Given any y in [
(e) If x
1 or x
(f) f x
1 for x
x
1
1], then y
f x
1
f x for some x in [ 1 1].
1, then f x
0 and f x
6. Consider the function f x (a) Let L
1
1 or f x
1 for x
x 2 2x 2
if x if x
1.
0.
1 1
lim f x . Find L .
x
0
(b) Find a number δ
0 so that if 0
(c) Show that f does not have a limit at
δ, then f x
x
L
0 01.
1.
(d) Explain what would go wrong if you tried to show that lim f x x
138
1
1 using the ε-δ definition.
HINT Try ε
1 2.
140
CHAPTER 2 SAMPLE EXAM
7. Let f be the function whose graph is given below. y 2 f 1
0
(a)
1
2
Sketch a plausible graph of f .
3
x
(b) Sketch a plausible graph of a function F such that F
f and F 0
1.
2
3
1
2 1 _1 _2
8. Suppose that the line tangent to the graph of y 4 1.
f x at x
(a) Find f 3 .
(b) Find f 3 .
(c) What is the equation of the line tangent to f at 3?
139
3 passes through the points
2 3 and
CHAPTER 2 LIMITS AND DERIVATIVES
9. Give examples of functions f x and g x with lim f x
lim g x
x
(a) lim
x
and
f x g x
x
(b) lim
f x
6
g x
x
(c) lim
f x
0
g x
x
(d) Is it possible to have lim x
f x
1? Either give an example or explain why it is not possible.
g x
10. Each of the following limits represent the derivative of a function f at some point a. State a formula for f and the value of the point a. (a) lim h
h
3
2
9
(b) lim
h
0
(c) lim x
3
x
1 x
x
3 2
3
8
2x
2
x
1
1
sin π 2
(d) lim h
0
140
h h
0
CHAPTER 2 SAMPLE EXAM
11. Let 3
x
x2
f x
27 x
(a) Evaluate each limit, if it exists. (i) lim f x (ii) lim f x x
1
(v) lim f x x
x
x
1 x 3
3
(iii) lim f x x
1
(vi) lim f x
3
if x if 1 if x
1
(vii) lim f x
3
x
9
(iv) lim f x x
3
(viii) lim f x x
6
(b) Where is f discontinuous?
12. The graph of f x is given below. For which value(s) of x is f x not differentiable? Justify your answer(s).
141
CHAPTER 2 LIMITS AND DERIVATIVES
13. A bicycle starts from rest and its distance travelled is recorded in the following table at one-second intervals. t (s) 0 1 2 3 4 5 6 d (ft) 0 10 24 42 63 84 5 107 (a) Estimate the speed after 2 seconds.
(b) Estimate the speed after 5 seconds.
(c) Estimate the speed after 6 seconds.
(d) Can we determine if the cyclist’s speed is constantly increasing? Explain. 14. Referring to the graphs given below, find each limit.
(a) lim x
0
(e) lim x
f x
(b) lim g x
g x
1
x
g x
f x
15. Draw a graph of f x
g x
(c) lim
1
(f) lim x
f x
x
x
f x
1
(g) lim x
1
ln ln x
(a) Over the range [2 10]. (b) Over the range [2 100]. (c) What is lim ln ln x ? x
142
1
f x
g x f x
(d) lim x g x x
2
SAMPLE EXAM SOLUTIONS
2
1. (a) lim f t t
, lim f t t
0
0
1, lim f t t
(b) lim f t exists for all x except x
0 and x
(c) There is a vertical asymptote at x
0.
t
x
(d) There is a horizontal asymptote at y (e) f is discontinuous at x 2. Solve 3 2
1
t
2.
1.
5a
b to get a
6, b
3. Taking lim f x gives a horizontal asymptote at y
a
x
asymptote at x lim f x 0. x
a
1
0, 2, and 4.
b and 52
2a
3, lim f t
2
The function is undefined at x
Algebraic simplification gives a vertical 0, but there is no asymptote there because
0
4. f x
x 4 4 x
x
1
(a) The domain is all values of x except x
(b) Algebraic simplification gives a limit of (c) f is not continuous at x be 15 . 5. (a) C. True for f x
1 and x
4.
1 5.
, for it is not defined there. It can be modified by defining f
x , untrue for f x
x2
x
1
x
1
x
1
to
(b) A. True by the Intermediate Value Theorem (c) C. True for f x
x , untrue for f x
x2
(d) A. True by the Intermediate Value Theorem (e) C. True for f x
x , untrue for f x
x2
(f) B. lim f x does not exist, contradicting the continuity of f . x
6. (a) L
0
0
(b) Let δ be any number greater than zero and less than
0 01 2 .
δ
0 07 works, for example.
(c) The left hand limit is 2, and the right hand limit is 1. 1 (d) Choose ε 21 . We now need a δ such that f x 1 1 δ But if x 2 for all x with x as x approaches 1, f x approaches 2, and f x 1 approaches 1, which is greater than 12.
7. (a) Answers will vary. Look for: (i) zeros at 1 and 2 143
1,
(ii) f positive for x
[0 1 and 2 3]
(iii) f negative for x (iv) f flattens out for x
1 2 25
145
CHAPTER 2 LIMITS AND DERIVATIVES
(b) Answers will vary. Look for (i) F 0
1
(ii) F always increasing (ii) F is never perfectly flat (iv) F is closest to being flat at x (v) F is concave up for x
0 1 and x
(vi) F is concave down for x 8. (a)
3
1 4
2 2 3
1 2
2 3
(b) The equation of the tangent line is y
3
2 3
x
2 , so f 3
(c) The equation of the tangent line is y
3
2 3
x
2.
2 3
3
2
3
1 3.
9. Answers will vary; the following are samples only. (a) f x
x 2, g x
x
(b) f x
6x , g x
x
(c) f x
x, g x
x2
(d) This is not possible. For lim
f x
g x contradicts the assumption that lim f x
1, either f or g would have to be negative for large x This
x
x
x 2, a
10. (a) f x (c) f x 11. (a) (i)
x 2
lim g x
3 1
3 2
(ii) 1
,a
3 (iii) Does not exist
(b) f is discontinuous at x
.
x
(b) f x
2x , a
(d) f x
sin πx , a
(iv) 9
(v) 9
1
(vi) 9
2 (vii) 3
(viii) 3
1.
12. f isn’t differentiable at x 1, because it is not continuous there; at x tangent there; and at x 4, because it has a cusp there.
2 because it has a vertical
13. (a) Answers will vary. One good answer would be to compute the average speed between 1 and 2 (14 ft/s) and the average speed between 2 and 3 (18 ft/s) and average them to get 16 ft/s. This is also the answer obtained by computing the average speed between 1 and 3. (b) Answers will vary. Using reasoning similar to the previous part, we get an estimate of 22 ft/s, but it could be argued that a number closer to 22 5 would be more accurate. (c) Answers will vary. The average speed between t
5 and t
6 is 22 5 ft/s
(d) Since we are given information only about the cyclist’s position at one-second intervals, we cannot determine if the speed is constantly increasing. 14. (a)
1 2
(b) 0
(c) Does not exist
144
(d)
4
(e) 1
(f) 2
(g) 0
CHAPTER 2 SAMPLE EXAM SOLUTIONS
15. (a)
(b)
y
(c) lim ln ln x
y
x
2
1
1 0
10
x 0
_1
100
_1
145
x
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