ch. 8 solutions new calc book

8/2/2019 Ch. 8 Solutions New Calc Book

1/28

350 Section 8.1

51. =y xsec2 , so the area is 2 1 22

0

4

tan sec , x x dx( ) + ( )which using NINT evaluates to 3 84. .

52.x=1 1

21

11

2 2yx

y y yand so the area is =

+

,

2

1

2dy,

which using NINT evaluates to 5.02.

53. (a) The two curves intersect atx = 1.2237831. Store thisvalue asA.

Area = ( sin sec ) . .2 1 3660

+ = x x dxA

(b) Volume = ( sin ) (sec ) . .2 16 4042 20

+ ( ) = x x dxA

(c) Volume = ( sin sec ) . .2 1 62920

+ = x x dxA

54. (a) Average temp

=

1

14 680 10

1287

6

14cos .

tdt F

(b) F tt

( ) cos=

80 1012

78

for

5 2308694 18 766913. . . tStore these two values asA andB.

(c) Cost =

0 05 80 10 12 78 5 10. cos .t

dtA

B

The cost was about $5.10.

55. (a)15600

24 1606004

29

17

( )t tdt

+ people.

(b)

1515600

24 160

1115600

24 16

29

17

2

( )

(

t t

dt

t t

++

+

00104 048

17

23

),dt

dollars

(c) = H E L( ) ( ) ( )17 17 17 380 people.H(17) is thenumber of people in the park at 5:00, and H ( )17 is therate at which the number of people in the park is

changing at 5:00.

(d) When = =H t E t L t ( ) ( ) ( ) ;0 that is, at t= 15.795

Chapter 8

Sequences, LHpitals Rule,

and Improper Integrals

Section 8.1 Sequences (pp. 435443)

Quick Review 8.1

1. f( )55

5 3

5

8= =

+

2. f( ) =

+= 2

2

2 32

3. + =2 3 1 1 5 1( )( . )

4. + =7 5 1 3 5( )( )

5. 1 5 2 124 1. ( ) =

6. =

2 1 5 4 53 1

( . ) .

7. lim limx x

x x

x x

x

x

+

+= =

5 2

3 16

5

30

3 2

4 2

3

4

8. limsin ( )

limx x

x

x

x

x = =

0 0

3 33

9. lim sin limx x

xx

xx

=

=1 1

1

10. lim .x

x x

x

x

x

++

= = 2

1

23 2 3Does not exist, or

Section 8.1 Exercises

1.1

2

2

3

3

4

4

5

5

6

6

7

50

51, , , , , ;

2. 25

2

8

3

11

4

14

5

17

6

149

50, , , , , ;

3. 29

4

64

27

625

256

7776

31252 48832, , , , . ,

117649

466562 521626

51

502 691588

50

. , .

4. 2 2 0 4 10 18 2350, , , , , ,

5. 3 1 1 3 11, , , ;

6. 2 1 0 1 5, , , ;

7. 2 4 8 16 256, , , ;

8. 10 11 12 1 13 31 19 487171 10 1 1 7, , . , . ; . ( . )

9. 1 1 2 3 21, , , ;

10. 3 2 1 1 2, , , ;

11. (a) 3

(b) a d+ = + =7 2 7 3 19( )

(c) a an n

= +1 3

(d) a n nn = + = 2 1 3 3 5( )( )

12. (a) 2

(b) a d+ = + =7 15 7 2 1( )(c) a a

n n= 1 2

(d) a n nn

= + = +15 1 2 2 17( )( )

13. (a)1

2

(b) a d+ = +

=7 1 71

2

9

2


2/28

Section 8.1 351

13. Continued

(c) a an n

= +11

2

(d) a nn

n = +

=1 11

2

1

2( )

( )+

14. (a) 0.1

(b) a d+ = + =7 3 7 0 1 3 7( . ) .

(c) a an n= +1 0 1.

(d) a n nn = + = +3 1 0 1 0 1 2 9( )( . ) . .

15.(a)1

2

(b) 81

20 03125

8

= .

(c) a an n=

1

21

(d) an

n

n=

= 81

22

1

4

16. (a) 1 5.

(b) ( )( . ) .1 1 5 25 62898

(c) a an n

= ( . )1 5 1

(d) an

n n= = ( )( . ) ( . )1 1 5 1 51 1

17. (a) 3

(b) ( ) ,

= 3 19 6839

(c) a an n= 3 1

(d) ann n= = ( )( ) ( )3 3 31

18. (a) 1

(b) ( )( )5 1 58 =

(c) a an n= 1

(d) ann= 5 1 1( )

19.7 2

33

=

( )

a1 2 3 5= =

a a nn n= + 1 3 2for all

20.

=3 5

42

a1 5 2 4 13= =( )( )

a n nn = + = +13 1 2 2 15( )( )

21. r =

=3010000

301010

1 3/

a1 33010

103 01= = .

a nnn= 3 01 10 11. ( ) ,

22. r =

=16

1 22

1 5

/

/

a11 2

21 4=

=/

/

a nnn n= ( ) ( ) ,1 2 11 3

23.

[0, 20] by [0, 1]

24.

[0, 20] by [1, 1]

25.

[0, 20] by [5, 5]

26.

[0, 20] by [0, 10]

27.

[0, 10] by [25, 200]

28.

[0, 10] by [5, 30]


3/28

352 Section 8.1

29.

[0, 20] by [1, 5]

30.

[0, 10] by [15, 10]

31. lim lim ( ) limn n n

n

n n

+= +

= =3 1

31

3 0 3

converges, 3

32. lim limn n

n

n

n

n += =

2

3

22

converges, 2

33. lim limn n

n n

n n

n

n

+ +

= =2 1

5 2

2

5

2

5

2

2

2

2

converges, 2 5/

34. lim lim limn n n

n

n

n

n n += = =

2 21

10

converges, 0

35. n kn

nn

n= +

=

2 11

31, l im ( )

n kn

nn

n= +

=

2 1 11

31, l im ( )

diverges

36. n kn

n

n

nn

n

n

n= ++

= =

2 11

11 0

2 2, lim ( ) lim ( )

n kn

n

n

nn

n

n

n= ++

= =

2 1 11

11

2 2, lim ( ) lim ( ) 00

converges, 0

37. lim ( . )n

n

= 1 1

diverges

38. lim ( . )n

n

=0 9 0

converges, 0

39. lim sin limn n

nn

nn

=

=

1 11

converges, 1

40. lim cosn

n

2

12

1cos ,n

diverges

41. limsin

n

n

n

1 1

n

n

n n

sin

lim limn nn n

=

=1 1

0

42. limn n

1

2

lim limn nn n

=

=1 1

0

Note:1

2for

n nn<

11 .

43. lim!n n

1

lim limn nn n

=

=1 1

0

Note:1

!for

n nn

11

44. limsin

n n

n

2

2

lim limn nn n

=

=1 1

0

Note:1

2for 1

n nn<

1

45. Graph (b)

46. Graph (c)

47.Table (d)

48. Table (a)

49. False. Consider the sequence with nth term

a nn = +5 2 1( ). Here

a a a a= = = =5 3 1 12 3, , , .and 4

50. True. aa

ar1

1

0 02> = >, , and

a a r nnn= >

11 0 for all 2.

51. C.5

=( )1

23

+ =1 3 5 14( )

52. E.1 25

2 5

1

2

.

.=

2 5

1 25

.

/=

53. D. lim sinn

nn

3

=

=

limn

nn

33


4/28

Section 8.2 353

54. E. n kn

nn

n=

+

=

2 13 1

21, lim ( )

n kn

nn

n=

+

=

2 1 13 1

21, lim ( )

55. (b)lim sinn

nn

2

=

=

limn

nn

2 2

56. (a) 1, 1, 2, 3, 5, 8, 13, 21, 34, 55

(b)

[0, 10] by [10, 60]

57. a arn

n= 1 implies that log a a n r n

= + log ( ) log .1 Thus

logan{ } is an arithmetic sequence with first term log a andcommon ratio log r.

58. a a n d n = + ( )1 implies that

10 10 10 101 1an a n d a d n= =+ ( ) ( ) . Thus 10an{ } is a geometric

sequence with first term10a and common ratio10d.

59. Given > 0 choose M =1

. Then

10

nn M < > if .

Section 8.2 LHpitals Rule (pp. 444452)

Exploration 1 Exploring LHpitals Rule

Graphically

1. limsin

limcos

x x

x

x

x

= =

0 0 11

2. The two graphs suggest that lim lim .x x

y

y

y

y =

01

20

1

2

3. yx x x

x5 2

=cos sin

.

The graphs ofy3 and y5 clearly show

that lHpitals Rule does not say that limx

y

y01

2

is equal to

lim .x

y

y

01

2

Quick Review 8.2

1.

x 10 1

+

.

x

x

1 1.1000

10 1.1046

100 1.1051

1000 1.1052

10,000 1.1052

1,000,000 1.1052

As xx

x

+

,.

10 1

approach 1.1052.

2. x xx1/(ln )

0.1 2.7183

0.01 2.7183

0.001 2.7183

0.0001 2.71830.00001 2.7183

As x x x +0 1, /(ln ) approaches 2.7183.

3.

x 11

x

x

1 0.5

0.1 0.78679

0.01 0.95490

0.001 0.99312

0.0001 0.99908

0.00001 0.99988

0.000001 0.99999

As xx

x

0 11

, approaches 1.

4.

x 11

+

x

x

1.1 13.981

1.01 105.77

1.001 1007.9

1.0001 10010

As x

x

x

+

1 11

, goes to .

5.

As tt

t

11

1, approaches 2.


5/28

354 Section 8.2

6.

As approaches 2.xx

x +

+,

4 1

1

2

7.

As approaches .xx

x 0

33,

sin

8.

As approaches1.

+2 2,

tan

tan

9. yh

h=1

sin

10. y hh= +( ) /1 1


1. limx

x

x

2 2

2

4

appears to be about

1

4;

[2, 4] by [1, 4]

By LHpitals Rule:

lim lim( )x x

x

x =

= =

2 2 2

2

4

1

2 2

1

4

2. limsin ( )

x

x

x

=

0

5appears to be about 5;

[2, 2] by [2, 6]

By LHpitals Rule:

limsin ( )

limcos( ( ))

x x

x

x

=

0 0

5 5 5 0

1 = 5

3. limx

x

x

+

2

2 2

2appears to be about

1

4;

[2, 4] by [1, 1]

By LHpitals Rule:

lim lim( )

x x

x

x

+

=

+

2 2

1 2

2 2

2

1

22 2

1

=1

4

4. limx

xx

1

3

11

appears to be about 13

;

[2, 2] by [1, 2]

By LHpitals Rule:

lim lim

( )

x x

x

x

=

1

3

1

2 3

1

2

1

31

1

=1

3

5. limcos

limsin

limx x

x

x

x

x

=

=0 2 0

1

2 xx

=0

0

2

1

2

cos( )

6. limsin

cos ( )lim

cos

/ /

+

=

2 2

1

1 2

=

2 2

2

4 22

2

sin( )

limsin

cos/

=1

4

7. limcos

limsin

t t t t

t

e t

t

e

=

0 0

1

1 1

=

=

limcos

t e0 00

1

8. lim limx x

x x

x x

x

x

+ +

=

2

2

3 2 2

4 4

12 16

2 4

3 112

2

6 2

1

62

=

=

lim( )x

9. (a) limsin

sinlim

cos( ( ))

cx x

x

x

=0 0

4

2

4 4 0

2 oos( ( ))2 02

=

(b) limsin

sinlim

cos( ( ))

cx x

x

x + +

=0 0

4

2

4 4 0

2 oos( ( ))2 02

=


6/28

Section 8.2 355

9. Continued

(b)

[2, 2] by [3, 3]

10. (a) limtan

limsec

x x

x

x

x

=

=

0 0

2

11

(b) limtan

limsec

x x

x

x

x

+ +

=

=

0 0

2

11

[1.5, 1.5] by [2, 10]

11. (a) limsin

limcos( )

( )x xx

=

0 3 0 2

0

3 0

x

=

(b) limsin

limcos( )

( )x x

x

x + +

=

0 3 0 20

3 0=

[2, 2] by [2, 10]

12. (a)lim

tan

lim

sec ( )

( )x x

x

x

=

0 2 0

2 0

2 0 =

(b) limtan

limsec ( )

( )x x

x

x + +

=+

02

0

2 0

2 0 =

[1, 1] by [10, 10]

13. Left:

limcsc

cotx

x

x

+

=

1

limcsc cot

cscx

x x

x

=

2

1

Right:

limcsc

cotx

x

x + +

= 1

limcsc cot

cscx

x x

x +

=

2

1

limit = 1

[3/4, 5/4] by [5, 5]

14. Left:

limsec

tan/x

x

x

+

= 2

1

limsec tan

sec/x

x x

x

=

22

1

Right:

limsec

tan/x

x

x +

+

= 2

1

limsec tan

sec/x

x x

x +

=

22

1

limit = 1

[/4, 3/4] by [5, 5]

15. limln( )

logx

x

x

+=

1

2

lim

ln

lnx

x

x

+

=

1

1

1

2

2

[0, 100] by [1, 2]

16. limx

x x

x

+

=

5 3

7 1

2

2

limx

x

x

=10 3

14

5

7

[0, 100] by [1, 2]

17. l im ( ln )x

x x +

= 0

0 i

limln

/lim

/

/l

x x

x

x

x

x + +

=

=0 0

21

1

1iim ( )

x

x +

=0

0


7/28

356 Section 8.2

18. lim tanx

xx

= 1

0i

lim tan ( ) limx hh

hh

h

=

=1 1

1

19. lim (csc co t cos )

x

x x x

+ + =

0

limcos cos sin

sinlim

sin

x x

x x x

x + +

+

=0 0

1 x x x x

x

+

=

sin cos

cos

2 2

1

20. lim (ln( ) ln( ))x

x x

+ = 2 1

limln( )

ln( )ln( )

x

x

x +

=2

12

21. lim ( ) ( )/x

x xe x

+ = + =0

1 1 0 1

limln( )

limln

lix

x

x

e x

x

x x

x

+

=+

=0 0

mmx

x

+

=0

11

12

lim l n ( )x

f xe e

=

0

2

22. lim( )( )x

xx

=1

1 1 1

limln

lim/

x x

x

x

x

= =

1 11

1

11

lim l n ( )x

f xe e e

= =

1

1

23. lim( ) ( ( ) )x

xx x

+ = + =1

2 1 2 1 1 02 1 1 2 1 1 0

lim ln( )/

lim /( )x x

x xx

xx

+ = =1

2

1 22 1

1 1

2

11 1

llim ( )x

xx

=

1

2

2 11

0

lim l n ( )x

f xe e

= =

1

0 1

24. lim (sin )x

xx

+=

0

00

limln(sin )

/lim tan

/x x

x

xx

x + +

=

0 0 2

1

1

1

=

=

+

+

limtan

limsec

x

x

x

x

x

x

0

2

02

2

= =

0

10

lim l n ( )

x

f xe e

+= =

0

0 1

25. lim ( )x

x

x ++

= + = 0

0 01

11

lim limln( )

/lim

x

x

x xx

x

x + + +

=+

=0 0

11

11

1 00 2

0

2

1

1

1

1

+

+

+

=+

x x

x

x

x xx

( )

/

lim( )

=+

= +lim

x

x

x0 10

lim l n ( )

x

f xe e

+= =

0

0 1

26. lim (ln ) /x

xx

= 1

lim ln lim /x x

xx

x

= =11

0

lim ln ( )x

f xe e

= =0 1

27. (a)x 10 102 103 104 105

f(x) 1.1513 0.2303 0.0345 0.00461 0.00058

Estimate the limit to be 0.

(b) limln

limln

lim/

x x x

x

x

x

x

x

= = = =

5 5 5

1

0

10

28. (a)x 100 10 1 10 2 10 3 10 4

f(x) 0.1585 0.1666 0.1667 0.1667 0.1667

Estimate the limit to be1

6.

(b) limsin

limcos

x x

x x

x

x

x + +

=

0

30

2

1

3

=

=

=

+

+

limsin

limcos

x

x

x

xx

0

0

6

61

6

29. Let f( )sin

sin

.

=3

4

100 10 1 10 2 10 3 10 4

f( ) 0.1865 0.7589 0.7501 0.7500 0.7500

Estimate the limit to be =3

4.

limsin

limcos

cos

= =

0 04

3 3

4 4

3

4

sin3

30. Let f tt

t t

t t( )

sin

sin= =

1 1sin

.t

t 100 10 1 10 2 10 3

f(t) 0.1884 0.0167 0.0017 0.00017


limsin

sin

sin

lim

t t

t

t t

t t

t t

=

=

0 0

1 1lim

+

= + +

0

0

1 cos

cos sin

limsin

sin cos c

t

t t t

t

t t tt oos t=0


8/28

Section 8.2 357

31. Let f x x x( ) ( ) ./= +1 1

x 10 102 103 104 105

f(x) 1.2710 1.0472 1.0069 1.0009 1.0001


l n ( )ln( )

limln( )

lim

f xx

x

x

x

x

x x

=+

+=

+=

1

1

1

1

1

0

110

1 1

=

+ = =

lim ( ) lim ( ) lim / ln ( )x

x

x x

f x x f x e == =e0 1

32. Let f xx x

x x( ) .=

+2

3 5

2

2

x 10 102 103 104 105

f(x) 0.5429 0.6525 0.6652 0.6665 0.6667

Estimate the limit to be 2

3.

lim lim lim x x x

x x

x x

x

x

+

=

+= =

2

3 5

1 4

6 5

4

6

2

2

2

3.

33. limsin

limcos

( )( )cos( )

= = =

0

2

0

222

12 0 0 0

34. limln sin

lim

cost t

t

t t

tt

=1 1

1 1

1

=

1

1 1( )

= +

1

1

35. limlog

log ( )lim

ln

( ) ln

x x

x

x

x

x

+=

+

2

3 3

1

2

1

3 3

=+

lim

( ) ln

lnx

x

x

3 3

2

=+

lim

ln ln

lnx

x

x

3 3 3

2

=

limln

lnx

3

2

= lnln

3

2

36. limln( )

lnlim

y y

y y

y

y

y y

y

+ +

+=

++

0

2

0

22

2 2

2

1

=+

+

=+

+

+

+

lim( )

lim( )

y

y

y y

y y

y y

y y

02

0

2

2

2 2

2

2 2

2

=++

+lim

y

y

y0

4 2

2 2

=++

= =4 0 2

2 0 2

2

21

( )

( )

37. lim tan lim

si

/ /y yy y

y

=

2 22

2 nn

cos

y

y

=

+

lim

cos ( ) sin

sin/y

y y y

y

2

21

=

+

2 2 21

2

2

cos ( ) sin

sin

=

=( )( )

( )

1 1

11

38. lim ( sin ) limsinx x

x xx

x + + =

0 0

1 1 1n n n

Let f xx

x( )

sin.=

limsin

lim

limx x

x

x

x x

+ += =

0 0

11

cosTherefore,

00 0

1 1 0+ +

= = =

(ln sin ) lim ln ( ) x x f xx

n ln

39. lim limx xx x

xx + +

= = 0 0

1 1 1

40. The limit leads to the indeterminate form 0.

Let f xx

x

( ) .=

12

ln ln

ln1 1

1

12 2

2

xx

x

x

x

x

=

=

lim

ln

lim

/

/

/li

x x

x

x

x

x

x

=

=0

2

0

3

2

2

1

1

2

1

1mm

xx

=02 0

lim lim l n ( )x

x

x

f x

xe e

= = =0 2 0

01

1

41. lim limx x

x

x x x

+

=

=3 5

2 2

3

4 10

2


9/28

358 Section 8.2

42. limsin

tanlim

cos

secx x

x

x

x

x = =

0 0 2

7

11

7 7

11 11

7

11


Let f x x x( ) ( ) . /( ln )= +1 2 1 2

ln( )

( )

ln

/( ln )

1 2

1 2

2

1 2

+ =

+x

x

x

x ln

limln( )

lnlim lim

x x x

x

x

x

x

x

+=

+=

+1 2

2

2

1 2

2 1 2xx x= =

lim

1

2

1

2

lim ( ) lim /( ln ) ln ( ) / x

x

x

f x x e e e

+ = = =1 2 1 2 1 2


Let f x x x( ) (cos ) .cos=

ln (cos ) (cos ) ln (cos )ln(cos )

sec

cos x x xx

x

x = =

limln(cos )

seclim

sincos

sec/ /x x

x

x

xx

=

2 2 xx xtan

=

lim

tan

sec tan/x

x

x x 2

= = lim cos

/x

x 2

0

lim (cos ) limcos ln ( )

x

x

x

f x x e e

= = = = =

2 2

0 1

45. The limit leads to the indeterminate form1.

Let f x x x( ) ( ) ./= +1 1

ln 1+ln 1+

1+

( )( )

limln( )

lim

1

/x

x

x

x

x

x x

1

0 0

=

= + +

x

11

11

0

1

0

1

+=

= = = + +

x

x e e ex

x

x

f xlim ( ) lim / ( )1+ ln

46. The limit leads to the indeterminate form

Let f x xx( ) (sin ) tan=

ln sin ln sinln sin

cot( ) tan ( )

( )

lim

tan x x xx

x

x

x

= =

00 0 2+ += = ln sin

cot

( )lim

cossin

csclim

x

x

xx

xx x 00

0 0

0+

+ +

==

( sin cos )

lim (sin ) limtan l

x x

x ex

x

x

nn ( )f x e= =0 1

47. The limit leads to the indeterminate form1.

Let f x x x( ) . /( )= 1 1

lnln /( )x

x

x

x1 1

1

=

limln

limx x

x

x

x

+ +=

=

1 11

1

11

lim lim /( ) ln ( )

x

x

x

f x x e ee

+ +

= = =1

1 1

1

1 1

48. dtt

t x x xxx

x

x

x

= = = ln ln ln ln22

2 2

lim lim ln lim ln lnx xx

x

x

dt

t

x

x = = =

2 22 2

49. lim lim / x x

x

x x

x

x

=

=1

3

3 1

2

2

1

4 3

3

12 13 11

50. lim lim lim x x x

x x

x x

x

x

++ +

=++

=2 3

1

4 3

3 1

4

6

2

3 2 xx= 0

51. lim

cos

limsin sin

limx

x

x

t dt

x

x

x

=

=1

1

2 1 21

1

1 xx

x

x=

1 2

1

2

cos cos

52. lim limln ln

lim/

x

x

x x

dt

t

x

x

x

=

=1

1

3 1 3 11

1

1

1 xx

x31 3

2= /

53. (a) LHpitals Rule does not help because applying

LHpitals Rule to this quotient essentially inverts

the problem by interchanging the numerator and

denominator (see below). It is still essentially the same

problem and one is no closer to a solution. Applying

LHpitals Rule a second time returns to the original

problem.

lim lim( / )( )

( / ) (

/

x x

x

x

x

x

+

+=

++

9 1

1

9 2 9 1

1 2

1 2

11

9 1

9 11 2

)lim

/ =

+

+xx

x

(b)

The limit appears to be 3.

(c) lim limx x

x

x

x

x

+

+=

+

+= =

9 1

1

91

11

9

13

54. (a) LHpitals Rule does not help because applying

LHpitals Rule to this quotient essentially inverts theproblem by interchanging the numerator and denominator

(see below). It is still essentially the same problem and

one is no closer to a solution. Applying LHpitals Rule

a second time returns to the original problem.

limsec

tanlim

sec tan

seclim

/ / x x x

x

x

x x

x = =

2 2 2 /

tan

sec2

x

x


10/28

Section 8.2 359

54. Continued

(b)


(c) limsec

tanlim

cos

sin

cos

lim/ / x x x

x

x

x

x

x

= =

2 2

1

=

/ sin2

11

x

55. Find c such that lim ( ) .x

f x c

=0

lim ( ) limsin

x xf x

x x

x =

0 0 3

9 3 3

5

=

lim

cos

x

x

x0 29 9 3

15

= limsin

x

x

x0

27 3

30

= = =

limcos

x

x

0

81 3

30

81

30

27

10

Thus, c =27

10. This works since lim ( ) ( ),

x f x c f

= =

00 so f is

continuous.

56. f x( ) is defined at xx

00

.lim f x( ) leads to the indeterminate

form 00.

ln lnln

ln

x x xx

x

x

x

x

x

x

x x

= =

=

=

1

1

1

10 02

lim lim limxx

x

x

x

x

x

x e e

=

= = =

0

0 0

0

0

1lim limln x

Thus, f has a removable discontinuity atx = 0. Extend thedefinition ofby letting (0) = 1.

57. (a) The limit leads to the indeterminate form 1.

Let (k) = 1+

r

k

kt

.

ln ln 1

ln 1

1 f k kt r

k

tr

k

k

( ) = =

+

lim limk k

tr

k

k

tr

k

r

k

+

=

+ln 1

1

12

1

2

1

k

=+

= =

limk

r t

r

k

r trt

11

lim limk

k t

k

kt

Ar

kA

r

k +

= +

0 0

1 1

= A ekf k

0 lim

( )ln

= A ert0

(b) Part (a) shows that as the number of compoundings per

year increases toward infinity, the limit of interest

compounded ktimes per year is interest compounded

continuously.

58. (a) For xf x

g x

= =01

11,

( )

( ).

lim( )

( )x

f x

g x

=0

1

lim( )

( )x

f x

g x

= =0

2

1

2

(b) This does not contradict LHpitals Rule since

lim ( )x

f x

=0

2 and limx0

g(x) = 1.

59. (a) A t e dx e ext x

tt( ) = =

= +

0 0 1

lim ( ) lim( ) limt t t t

A t ee

= + = +

1 11

1= 1

(b) V t e dxxt( ) ( )= 20

= e dxxt 2

0

=

1

2

2

0

e xt

= +

1

2

1

2

2e

t

= +( )2

12e t

lim( )

( )lim

( ) (

t t

t

t

V t

A t

e

e

= +

+=

21

1

212 ))

1 2=

(c) lim( )

( )

lim( )

t t

t

t

V t

A t

e

e

+ +=

+

+0 0

2

21

1

=

+lim

( )

t

t

t

e

e0

2

22

= =

2

2

1

( )


11/28

360 Section 8.2

60. (a) x (x)

0.1 0.04542

0.01 0.00495

0.00 0.00050

0.0001 0.00005


(b) limsin

x

x

x += =

0 1 2

0

10

LHpitals Rule is not applied here because the limit is

not of the form0

0or

, since the denominator has

limit 1.

61. (a) f x ex x( ) ( )= ln 1+1/

11

0+ >x

when x x< >1 0or

Domain: ( , ) ( , ) 1 0

(b) The form is 0 1 , so lim ( )x

f x

= 1

(c) limx

x

ln 11

11

1+

=+

x

x

x

xlim

=

+

limx

x x

x

11

1

1

2

1

2

=

+

=lim

x

x

1

1

11

lim ( ) lim ( / )x x

x x f x e e

= =ln 1+1

62. False. Need g (a) 0. Consider f x x( ) sin= 2 and g(x) = x2

with a = 0. Here lim ( ) lim ( ) .x x

f x g x

= =0 0

0

63. False. The limit is 1.

64. C. limtan secx

x

x x= = =

0 2

1 1

11

65. D. lim lim lim /

x x x

x

x

x

x

x

x

= = =1

21

2

31

3

2

11

1 1

1

2 2

1 22

66. B. limlog

loglim ln

ln

ln

lnx x

x

x

x

x

= =2

3

1

21

3

3

2

67. E. lim limln

/x

x

xxx

x

+

= +

1

1 11

1 3

3

= +

lim

( )

/x

x x

x

1

1

1 3 2

=+

= =

lim( )

limx x

x

x x

3

1

3

13

2

lim l n ( )x

f xe e

= 3

68. Possible answers:

(a) f x x g x x( ) ( ); ( )= = 7 3 3

lim( )

( )lim

( )lim

x x x

f x

g x

x

x =

=3 3 3

7 3

3

7

17

(b) f x x g x x( ) ( ) ; ( )= = 3 32

lim( )

( )lim

( )lim

( )

x x x

f x

g x

x

x

x

=

3 3

2

3

3

3

2 2 3

110=

(c) f x x g x x( ) ; ( ) ( )= = 3 3 3

lim ( )( )

lim( )

lim( x x x

f xg x

xx x

=

=3 3 3 3

33

13 3))2

=

69. Answers may vary.

(a) f x x g x x( ) ; ( )= + =3 1

lim( )

( )lim lim

x x x

f x

g x

x

x =

+= =

3 1 3

13

(b) f x x g x x( ) ; ( )= + =1 2

lim( )

( )lim lim

x x x

f x

g x

x

x x =

+= =

1 1

20

2

(c) f x x g x x( ) ; ( )= = +2 1

lim( )

( )lim lim

x x x

f x

g x

x

x

x

=

+= =

2

1

2

1

70. (a) Because the difference in the numerator is so small

compared to the values being subtracted, any calculator

or computer with limited precision will give the

incorrect result that1 6 cosx is 0 for even moderatelysmall values ofx. For example,

at x x= 0 1 0 99999999999956. , cos . (13 places), so on

a 10-place calculator, cosx6 1= and1 06 =cos .x

(b) Same reason as in part (a) applies.

(c) limcos

limsin

x x

x

x

x x

x =

0

6

12 0

5 6

11

1 6

12

=

limsin

x

x

x0

6

62

=

limcos

x

x x

x0

5 6

5

6

12

= =

limcos

x

x

0

6

2

1

2


12/28

Section 8.3 361

70. Continued

(d) The graph and/or table on a grapher show the value of

the function to be 0 forx-values moderately close to 0,

but the limit is 1/2. The calculator is giving unreliable

information because there is significant round-off error

in computing values of this function on a limited

precision device.

71. (a) = = f x x g x x( ) , ( )3 2 12

f f g g

c

c

c

( ) ( ) , ( ) ( )1 1 2 1 1 2

3

2 1

2

2

3 2

2

2

= =

=

= cc

c c

c c

++ =

+ =

1

3 2 1 0

3 1 1 0

2

( )( )

c c= = 1

31or

The value ofc that satisfies the property is c=1

3.

(b) f x x g x x = =( ) sin , ( ) cos

f f g g

20 1

20 1

=

=( ) , ( )

=

sincos

c

c

1

1

tan c = 1

c = =tan 114

on 0

2,

72. (a) ln f x g x f xg x( ) ( ) ( )( ) = ln

lim ( ( ) ( )) lim ( ) lim ( ) x c x c x c

g x f x g x f x

=

( )ln ln

(( )= = ( )lim ( ) lim( ) ( )

( )

x c

g x

x c

f xg x

f x e e

= = =ln 0

(b) lim ( ) ( ) lim ( ) lim ( ) x c x c x c

g x f x g x f x

( ) = ( )ln ln(( )= = ( )( )

lim ( ) lim( ) ( )( )

x c

g x

x c

f x g x f x e e

= = = ln

Quick Quiz Sections 8.1 and 8.2

1. C. lim ( ) ( / )/

xx x

x+

0

4 3

21 4 3 1

=+

lim

/ ( ) ( / )/

x

x

x0

1 34 3 1 4 3

2

=+

lim/ ( ) /

x

x

0

2 34 9 1

2

=2

9

2. D. lim ( )x

xx

+0

23

= +lim

ln

/x

x

x0 1 2

= = +lim

/

/x

x

x02

1

1 20

lim l n ( )x

f xe e = =0

03 3 3

3. B. limsin

x

xt dt

x

2

2

2 4

=

limcos cos

x

x

x2 22

4

= =

limsin sin

x

x

x2 2

2

4

4. (a)1 2

4

1

2

1 3/

/

=

=4

1 28

/

(b) 1

2

(c) an

n

n n=

= 81

21 21 4( ) ( )

(d) a an n=

1

2 1

Section 8.3 Relative Rates of Growth

(pp. 453458)

Exploration 1 Comparing Rates of Growth

as x

1. lim lim(ln )( )

lim(ln )

x

x

x

x

x

xa

x

a a

x

a a

= =

2

2

2 22= , so ax grows

faster than x2 as x .

2. lim lim .x

x

x x

x

= =

3

21 5

3. lim limx

x

x x

xa

b

a

b =

= becausea

b> 1.

Quick Review 8.3

1. limln

limx x x

x

x

x

e e = =

1

0

2. lim lim lim limx

x

x

x

x

x

x

xe

x

e

x

e

x

e

= = =

3 23 6 6==


13/28

362 Section 8.3

3. limx x

x

e=

2

2

4. lim lim limx x x x x x

x

e

x

e e = = =

2

2 2 2

2

2

2

40

5. 3 4x

6.2

23

2x

xx=

7. lim( )

( )lim

lnlim

x x x

f x

g x

x x

x

x

=

+=

+=

11

11

8. lim( )

( )lim lim

x x x

f x

g x

x x

x x =

+= + =

4 5

21

5

41

2

9. (a) f xe x

e

x

e

x

x x( ) =

+= +

2 2

1

=

=

f x

xe x e

e

x x

e

x x

x x( )

2 22

2

2

20

2x x

ex

=

x x( )2 0 =x x

f x x x

= = < < >

0 2

0 0 2

or

for or( )

The graph decreases, increases, and then decreases.

f fe

( ) ; ( ) .0 1 2 14

1 5412

= = +

fhas a local maximum at (2, 1.541) and has a localminimum at (0, 1).

(b)fis increasing on [0, 2](c)fis decreasing on (, 0] and [2, ).

10. f xx x

x

x

xx( )

sin sin,=

+= + 1 0

Observe thatsin

.x

xx x< < 1 0since sin forx

lim ( ) limsin

x xf x

x

x = + = + =

0 01 1 1 2

Thus the values offget close to 2 as x gets close to 0, sof

doesnt have an absolute maximum value.fis not defined

at 0.


1. lim lim lim lix

x

x

x

x

x

x

e

x x

e

x

e

+=

= =

3 23 1 3 3 6

mmx

xe

=

6

2. lim lim!x

x

x

xe

x

e

= =

20 20

3. lim limsin

, cos ,cos cosx

x

x x

x

x

e

e

e

x ex

=

1 1

llimsinx

x

y

e

x e =

cos

4. lim( / ) !( / )x

x

x

xe e

x = = 5 2 5 2

5. limln

lnlim

/

/lim

x x x

x

x x

x

x x x =

=

=

1

1

1

10

2

6. limln

lim/

/ ( ) ( )/ /x x

x

x

x

x x x = = =

1

1 2

1

1 21 2 1 200

7. limln

lim/

/ ( )lim

/ (/ x x x

x

x

x

x

= =3 2 3

1

1 3

1

1 3 xx x) /=

2 30

8. limln

lim/

lim x x x

x

x

x

x x = = =

3 2 3

1

3

1

30


x xx

xx

+ = + = =2

24 2 4

222

1


x x

x

x x

x

x

x

+=

+= =

4

2

4

4

2

2

5 5 12

1211

11. lim( )

lim lim/

x x x

x x

x

x x

x

x

+=

+=

6 2 1 3

2

6 2

6

120 33

31201

x=

12. limsin

limcos

, cos , lx x

x x

x

x x

xx

+=

+

2

2

2

21 1 iim

x

x

x=

2

21

13. lim logln

ln

/ lnxx

x

x

x = =

1

1

2

10

11

2 10

14. limx

x

x

e

ee

+

=1

15. First observe that 1 4+x grows at the same rate as x2.

lim lim lim x x x

x

x

x

x x

+= =

+= + =

1 1 11 1

4

2

4

4 4

Next compare x2 with ex .

lim lim limx x x x x x

x

e

x

e e = = =

2 2 20

x e x xx2 41grows slower than as so +,

grows slower than ase xx .

16. lim lim .x

x

x x

x

e e e =

= >4 4 4

1since

4x grows faster than ex asx .


14/28

Section 8.3 363

17. limln

lim

ln

x x x x

x x x

e

xx

x

e

=

+ 1

1

=lim

ln

x x

x

e

= =lim/

x x

x

e

1

0

x lnx x grows slower than ex asx .

18. lim limx

x

x x

xe

ex

= =

xex grows faster than ex asx .

19. limx x

x

e=

1000

0 Repeatedapplicationof L'Hpitaal's

Rulegets

=

lim!

.x xe

10000

x1000 grows slower than ex asx .

20. lim( ) /

limx

x x

x x x

e e

e e

+= +

=2 1

2

1

2

1

22

e ex x+

2grows at the same rate as e xx as .

21. lim limx x

x

xx

x

+= +

= 3

2 2

3 3

x3 + 3 grows faster than x2 asx .

22. lim limx x

x

x x x

+= +

=15 3 15 3

02 2

15x + 3 grows slower than x2

asx .

23. limln

lim/

lim x x x

x

x

x

x x = = =

2 2

1

2

1

20

lnx grows slower than x2 asx .

24. lim lim(ln )

lim(ln )

x

x

x

x

x

x

x x = = =

2 2 2

2

2 2

22

2

.

2x grows faster than x2 asx .

25. limlog

lnlim

log

lnlim

(ln

x x x

x

x

x

x

x

= =2

222 2 )) / ( ln )

ln ln

2 2

2x=

log22x grows at the same rate as lnx asx .

26. lim/

lnlim

lnx x

x

x x x = =

1 10

1

xx xgrows slower than asln .

27. limln

limlnx

x

x x

e

x e x

= =

10

e x xx grows slower than asln .

28. limln

lnx

x

x=

55

5lnx grows at the same rate as ln ,x xas

29. Compare e xx xto .

lim limx

x

x x

xe

x

e

x =

= 0

ex grows slower than xx.

Compare ex to (ln ) .x x

lim(ln )

limlnx

x

x x

xe

x

e

x =

= 0

e

x

grows slower than (ln ) .x

x

Compare x ex to ex/ .2

lim lim/

/

x

x

x x

xe

ee

= =

2

2

e ex xgrows faster than / .2

Compare xx to (ln ) .x x

lim(ln )

limlnx

x

x x

x

x

x

x

x

x =

= since lim

= = x

x

x

xxln /.lim

1

xx grows faster than (ln ) .x x

Thus, in order from slowest-growing to fastest-growing, we

get ex/2 , ex , (ln ) ,x x xx .

30. Compare 2x to x2

lim lim(ln )

lim(ln )

x

x

x

x

x

x

x x = = =

2 2 2

2

2 2

22

2

2 2x xgrowsfaster than .

Compare to (ln2)2x x .

lim(ln )

limlnx

x

x x

x

=

= 2

2

2

2since

2

ln 221> .

2x grows faster than (ln )2 x

Compare 2x to ex .

lim lim .x

x

x x

x

e e e =

= 1, is (ln ) .a an x We can

apply L Hpitals Rule n times to find lim .x

x

n

a

x

lim lim(ln )

!x

x

n x

n xa

x

a a

n = = =


16/28

Section 8.3 365

39. Continued

(b) Thus ax grows faster than x n asx for any positiveinteger n.

40. (a) Apply LHpitals Rule n times to find

lim .

x

x

n

n

n

n

e

a x a x a x a

+ + + +11

1 0

lim lim!x

x

nn

nn x

x

n

e

a x a x a x a

e

a n + + + +

=1

11 0

==

Thus ex grows faster than

a x a x a x an

nn

n+ + + +

11

1 0 as x .

(b) Apply LHpitals Rule n times to find

lim .x

x

nn

nn

a

a x a x a x a + + + +

11

1 0

limx

x

n

n

n

n

a

a x a x a x a

+ + + +=

1

1

1 0

= =

lim(ln )

!x

n x

n

a a

a n

Thus ax

grows faster than

a x a x a x an

nn

n+ + + +

11

1 0 asx .

41. (a) limln

lim lim/

( / )/x n x n x

x

x

x

nx

n

x = =

11 1

1

1

1 nn= 0

Thus lnx grows slower than x n1/ asx for anypositive integer n.

(b) limln

lim limx a x a x a

x

x

x

ax ax

= = =

11

01

Thus lnx grows slower than x a asx for anynumber a > 0.

42. limln

xn

nn

n

x

a x a x a x a + + + +

11

1 0

=+ + +

=

lim( )

lim

xn

nn

n

x

x

na x n a x a

1

1

1

11

21

nna x n a x a xnn

n

n

+ + +

=

( )1

0

1

1

1

Thus lnx grows than any nonconstant polynomial as to

x .

43. Compare n n nlog23 2to as n .

limlog

limlog

lim

(ln )

/n n

n

n n

n

n

nn

=

=

2

3 2

2

1 2

((ln )

limln

lim(ln

/

/

/

2

1

2

1

22

1 2

1 2

1 2

n

n

n

n

n

n

=

=

220

)=

Thus n nlog2 grows slower than n3 2 as n .

Compare n nlog2 to n n( log )22

limlog

(log )lim

logn n

n n

n n n = =2

22

2

10

Thus n nlog2 grows slower than n n(log )22 as n .

The algorithm of order ofn nlog2 is likely the most efficientbecause of the three functions, it grows the most slowly as

n .

44. (a) It might take 1,000,000 searches if it is the last item in

the search.

(b) log , , . ;2 1 000 000 19 9 it might take 20 binary searches.

45. (a) The limit will be the ratio of the leading coefficients of

the polynomials since the polynomials must have the

same degree.

(b) By the same reason as in (a), the limit will be the ratio

of the leading coefficients of the polynomial.

46. True. because limlog

/n

n n

n =2

3 2 0

47. False. They grow at the same rate.

48. E. lim lim!

!x x

x

x x

x

++ +

= = 6

5 2

1

1

6

5

49. A. limlog

x x x

x

e

x

e =

=3

1 130

ln

50. C. limx

x

x

e

ee

+

=2

2

51. D. lim lim lim x x x

x x

x

x x

x

x

+=

+=

8 4

4

8 4

8

36720

672001

3x=

52. (a) lim limx x

x

xx

= =

5

2

3

x5 grows faster thanx2.

(b) lim limx x

x

x = =

5

2

5

2

5

2

3

3

5 3x and2 3x have the same rate of growth.


17/28

366 Section 8.4

52. Continued

(c) m > n since lim lim .x

m

n x

m nx

xx

= =

(d) m = n since lim limx

m

n x

m nx

xx

= is nonzero and finite.

(e) Degree ofg > degree off(m > n) since lim ( )

( ).

x

g x

f x=

(f ) Degree ofg = degree off(m = n) since lim( )

( )x

g x

f xis

nonzero and finite.

53. (a) lim( )

( )lim

( )

( )lim

( )

x x x

f x

g x

f x

g x

f x

g =

=(( )x

=

Thus f grows faster than g asx by definition.

(b) lim( )

( )lim

( )

( )lim

( )

x x x

f x

g x

f x

g x

f x

g =

=(( )x

L=

Thus f grows at the same rate as g asx by

definition.

54. (a) lim( )

( )lim

( )

( )x x

f x

g x

f x

g x

= =

Thus f x( ) grows faster than g x( ) by definition.

(b) lim( )

( )lim

( )

( )x x

f x

g x

f x

g xL

= =

Thus f x( ) grows at the same rate as g x( ) bydefinition.

Section 8.4 Improper Integrals

(pp. 459470)

Exploration 1 Investigatingdx

xp0

1

1. Because1

xp

has an infinite discontinuity atx = 0.

2.dx

x

dx

xx

c cc c

c

= = = + +

lim lim ln lim0 0

1

0

1 1

00 + = ( l n )c

3. Ifp>1, then

dx

x

dx

xp

cpc

= +

lim0

1

0

1

= +

+

+limc

p

c

x

p0

1

1

1

= +

=

+

+

limc

pc

p0

11

1because (p + 1) < 0.

4. If0 1< 0 for 3 0 forx > 1

The domain is (1, ).

7. 1 cosx 1, socosx 1.cos cosx

x

x

x x2 22

1=

8. x2 1 x2 so x x x2 21 = forx >1

1

1

1

2x x

9. lim( )

( )lim lim

x x

x

x x

xf x

g x

e

e

e

e =

+

=4 5

3 7

4

3xx x

= =lim

4

3

4

3

Thusfand g grow at the same rate asx .

10. lim( )

( )lim

x x

f x

g x

x

x =

+

2 1

3

=

+limxx

x

2 1

3

=

+=

lim

x

x

x

21

13

2


18/28

Section 8.4 367


1. (a)2

1

2

12 200

x

xdx

x

xdx

b

b

+=

+

lim

(b) lim lim ln( )b

b

b

bx

xdx x

+= +( ) =

2

11

20

2

0

diverges

2. (a)dx

x

dx

xb

b

1 3 1 311=

lim

(b) lim limb b

bb dx

xx

=

= 1 3 2 31

1

3

2

diverges

3. (a) 2

1

2

12 2 2 2

0x

xdx

x

xdx

b b

b

( )lim

( )

lim

+=

++

+2

12 20

x

xdx

b

( )

(b) lim( )

lim( )b b b

bxx

dx xx

dx +

++

21

212 2

02 20

=

+

+

+

lim limb

bb

b

x x

1

1

1

12

0

2

0

== 0

converges

4. (a)dx

x

dx

xb

b=

lim 11

(b) lim lim ( )b b

bb dx

xx

= = 2

11

diverges

5.dx

x

dx

x xb b

bb

4 4 31

11

1

3

1

3= =

=

lim lim

6.2 2 2

21

3 3 21

11

dx

x

dx

x xb b

bb

= =

= lim lim

7.dx

x

dx

xx

b b

bb

3 3

2 3

111

3

2= =

lim lim ( ) ==

diverges

8.dx

x

dx

xx

b b

bb

4 4

3 4

111

4

3= =

= lim lim ( )

diverges

9.dx

x

dx

x xb bb

b2 2

11 1

1= =

=

lim lim

1

10.dx

x

dx

x xb b( )lim

( )lim

( )=

=

2 21

2 23 3 2

=

bb

000

1 8/

11.2

1

2

1

1

12 2dx

x

dx

x

x

xb b=

=

+

lim lim ln

=

b

b

222

3ln

12.3 3

31

2 3

dx

x x

dx

x x

x

xb b=

=

lim lim ln

=

222

3 2

bb

ln

13.dx

x x

dx

x x

x

xb b2 25 6 5 6

3

2+ +=

+ +=

++

lim lim ln

=

1

11

2

bb

ln

14. 2

4 3

2

4 32 2

00 dx

x x

dx

x xb b +=

+ lim

=

=

lim ln ln

b

b

x

x

1

33

0

15.5 6

2

5 6

22 211

x

x xdx

x

x xdx

b

b++

=+

+

lim

= +( ) =

lim ln (( ) ( ) )b

b

x x2 2 3

1

diverges

16.2

2

2

222 22

dx

x x

dx

x xb

b

b

=

=

lim

lim lnxx

x

b

=

22

2

ln

17. x e dx x e dx xxb

xb

b

= = 21 21 2 14lim lim = e ex2 23

4

18. x e dx x e dx

x x

x

b

x

b

b

2 200

2

2 2

=

=

lim

lim11

422

0

=

e

x

b

19. x x dx x x dxx x

b bln lim ln lim ln= =

2 2

12

24

bbb

11

=

diverges

20. ( ) lim ( )

lim (

x e dx x e dx

x

x

b

xb

b

+ = +

=

1 1

0 0

( ) =2 20

)e xb

21. e dx e dx e dxx

b

x

b

xb

b

= + =1 1 1 1 1 100

2lim lim


19/28

368 Section 8.4

22. 2 2 22 2 2

0 x e dx x e dx x e dx

x

b

x

b

xb

b

= + =lim lim00

lim limb

x

b b

xb

e e

( ) + ( ) =2 2

0

0

0

23.dx

e e

dx

e e

dx

e ex x b x x b x x

b

+=

++

+ lim lim 0bb0

lim tan ( ) lim tan ( )b

x

b b

xb

e e

( ) + ( ) =10

1

0 2

24. e dx e dx e dxx

b

x

b

xb

b

2 2 2

0

0= +

lim lim

=

+

=

lim lim

b

x

bb

xb

e e2

02

02 2

divverges

25. (a) The integral has an infinite discontinuity at the interior

pointx = 1.

(b)dx

x

dx

x

dx

xb b b

b

1 1 12

12

12

2

00

2

=

+

+ lim lim

=+

+

+lim ln lim l

b

b

b

x

x10

1

1

2

1

1

1

2nn

x

xb

+

=

1

1

2

diverges

26. (a) The integral has an infinite discontinuity at the endpoint

x = 1.

(b)dx

x

dx

xx

b b

bb

1 1 22 1 2 1

1

00=

= ( ) =

lim lim sin ( )

01


x = 0.

(b)x

x xdx

x

x xdx

x

b b

b

++

= ++

= +

12

1

22 0 2

1

0

1

0

2

lim

lim 22 31

xb

( ) =


x = 0.

(b)e

xdx

e

xdx e e

x

b

x

b

x

bb

= = ( ) = lim lim0 0

42

2 2 244

0

4


x = 0.

(b) x x dx x x dx

xx

b b

b

ln ( ) lim ln( )

lim ln

=

=

01

0

1

0

2

2

xx

b

21

41 4

= /

30. (a) The integral has an infinite discontinuity at the interior

pointx = 0.

(b)dx

x

dx

x

dx

xb b b

b

1 1 1 1 1 10 0

4

11

4= +

=

+ lim lim

llim ( )

lim ( )

b

b

b b

x x

x x

+

( ) +

( )

0 1

0

4

2

2

sign

sign == 6

31. 01

1

11

1

1

+ )

e e edx

x x xon convergesbecause, ,

converges.

32. 01

1

11

13 3 31

+

)

x x xdxon convergesbecause, ,

converges.

33. 01 2 1

+

)

xx

x xdx

cos, ,on divergesbecause

diverges.

34.dx

x

dx

x

dx

x

dx

x4 40 4 400

1

12

12

12

1+=

+=

++

+

aand on01

1

11

4 2

+ )

x x,

converges because121 x

dx converges.

35. y e dy y e dy ey

b b

y

b

x

b

= = ( )2 1

0

22 1

0

1lim limln lnnln 2

0

2

=

diverges

36.dr

r

dr

rr

b b

bb

4 42 4 4

4 4 000

4

=

= ( ) =

lim lim

37.ds

s s

ds

s s

s

b b b( )

lim

( )

lim tan

1 1

2

0 0

1

+

=

+

= ( ) =

bb

0

38.du

u u

du

u ux

b b b2 1 2

2

1

2

1

1 2

1 11

=

=

lim lim tan(( ) =b

2

3

39.16

1

16

1

1

2

1

200

tanlim

tanl

+=

+=

v

vdv

v

vdv

b

biim (tan )

b

b

v

( )=

8

2

1 2

0

2

40.

e d e d x eb b b

x= = ( ) lim lim ( )

00

0

1bb

= 1

41.dt

t

dt

t

dt

tb

b

b b1 1 11 00

2

1

2

=

+

=

+ lim lim limm ln( )

lim ln( )

b

b

b b

x

x

+

( )

+ ( ) =

1 0

1

2

1

1

divverges

42. ln(| |) lim ln(| |) lim lw dw w dwb

b

b

= + +

0 11

1

0

nn(| |)w dwb

1

= ( ) + ( )

+lim ln(| |) lim ln(| |)

b

b

b

w w w w w w0 1 0 bb

1

2=


20/28

Section 8.4 369

43. For x y 0 0, on [1,).

Area = =

lnx

xdx

x

xdx

b

b

2 211lim

ln

Integratelnx

xdx

2 by parts.

u x dv

dx

x

dux

dx vx

= =

= =

ln 2

1 1

ln ln lnx

x

x

x

dx

x

x

x xC

2 2

1= + = +

Area =

= +

lim

lnlim

ln

b

b

b

x

x x

b

b b

1 11

1

= 1

Notethat limln

lim/

.b b

b

b

b

= =

1

10

44. Forx 0 ,y 0 on [1, ).

Area = =

ln lim lnxx dx xx dxbb

1 1

Integrate by letting soln

.x

xdx u du

dx

x= =

lnx

xdx u du u C x C = = + = +

1

2

1

2

2 2(ln )

Area =

= =

lim (ln ) lim (ln )b

b

bx b

1

2

1

2

2

1

2

45. (a) Since f is even, f x f x( ) ( ). = Let u = x, du = dx.

f x dx f x dx f x dx( ) ( ) ( )

= + 00

= +

f u du f x dx( )( ) ( )1

0

0

= +

f x du f x dx( ) ( )00=

2 0 f x dx( )

(b) Since f is odd, f x( ). = f x( ). Let u = x,du = dx

f x dx f x dx f x dx( ) ( ) ( )

= +0

0

= +

f u du f x dx0

01( )( ) ( )

= + =

f u du f x dx( ) ( )00 0

46. (a) 2

1

2

12 200xdx

x

xdx

xb

b

+=

+ lim

= +

= + =

lim ln( )

lim ln ( )

b

b

b

x

b

2

02

1

1

Thus the integral diverges.

(b) Both2

1

2

12 2

0

0

xdx

x

xdx

x+ +

and must converge in order

for2

12xdx

x +

to converge.

(c) lim lim ln( )b b b

b

b

b x dx

x

x

+= +

2

1

12

2

= + +

= =

lim ln( ) ln( )

lim .b

b

b b2 21 1

0 0

Note that2

12x

x +is an odd function so

2

10

2

xdx

xb

b

+=

.

(d) Because the determination of convergence is not made

using the method in part (c). In order for the integral to

converge, there must be finite areas both directions

(toward and toward ). In this case, there areinfinite areas in both directions, but when one computes

the intergral over an interval [b, b], there is

cancellation which gives 0 as the result.47. By symmetry, find the perimeter of one side, say for

0 1 0 x y, .

y x2 3 2 31=

y x= ( )1 2 3 3 2

dy

dx x x x x=

= 3

21

2

312 3 1 2 1 3 1 3 2 3 1( ) ( ) 22

dy

dx x x x

= = 2

2 3 2 3 2 31 1( ) ( )

12

2 3 1 3+

= = dy

dxx x

x dx x dxb b

= +1 3

0

1

0

1 31

lim

=

=

+

+

lim

lim

b b

b

x

b

0

2 3

1

0

2 3

3

2

3

2

3

2==

3

2

Thus, the perimeter is 43

26

= .

48. False. See Theorem 6.

49. True. See Theorem 6.

50. C.dx

x

dx

x xb

b

b1 011 1 011 0 01

100. . .

lim lim

= =

=1

100

b


21/28

370 Section 8.4

51. B.dx

x

dx

xx

b b b b0 50

1

0 0 5

1

0

1

2 2. .

lim lim = = ( ) =

52. E.dx

x

dx

xx

b

b

b

b

=

= ( ) 1 1 10

1

1 0 1 0

lim lim ln(| |) ==

53. C.dx

x

dx

xx

b

b

b

b

20 201

01 1+ = + = ( ) =

lim lim tan

2

54. (a)dx

x

dx

xx

b

b

b

b

0 51 0 51 1

2. .

lim lim ,

= = ( ) = or it diverges.

(b)dx

x

dx

xx

b

b

b

b

1 1 1

= = ( ) = lim lim ln , or it diverges.

(c)dx

x

dx

x xb

b

b

b

1 51 1 511

2. .

lim lim

= =

== 2

(d)dx

x

dx

xp b p

b

1 1

= lim

=

=

lim

lim

b p

b

b p

p x

p b p

1

1

1

1

1

1 1

1

1

1

=

1

1

1

1

1

11

1

1

p

b pp blim

(e)dx

xp

p11

>,

=

limb pp b p

1

1

1 1

11

=

11

1 11p p

=1

1p

dx

xp

p11

1 large enough. Also we can

make f x d xb

( )

and f x d xb

( )

as small as we want bychoosing b large enough. This is because

0 12< < > f x e xx( ) .(/ for Likewise,

0 12< <


22/28

Section 8.4 371

57. (a) For x x x e ex x 6 62 6, ,so2

e dx e dxx x

2

6

6

6

=

limb

xbe dx6

6

= lim

bx

b

e16

6

6

= +

limb

be e

1

6

1

6

6 36

= < 1

64 10

36 17e

(b) e dx e dx e dx x x x = +

2

1

2

1

6 2

6

+ e dxx2

1

6 174 10

Thus, from part (a) we have shown that the error is

bounded by 4 10 17 .

(c) e dx xx x

2

11 6 0 1394027926NINT(e

2, , . ) .

(This agrees with Figure 8.16.)

(d) e dx e dx e dx x x x = +

2

0

2

0

3 2

3

+

e dx e dxx x2

0

3 3

3

since for >3. x x x2 3

e dx e dxx

b

xb

=333

3lim

=

limb

x

b

e1

3

3

3

= +

limb

be e

1

3

1

3

3 9

= a,

f x g x dxa

b

a

b( ) ( ) .

If the infinite integral ofg converges, then taking the

limit in the above inequality as b shows that theinfinite integral offis bounded above by the infinite

integral integral ofg. Therefore, the infinite integral off

must be finite and it converges. If the infinite integral off

diverges, it must grow to infinity. So taking the limit in the

above inequality as b shows that the infinite integralofg must also diverage to infinity.

59. (a) For n = 0:

e dx e dxx

b

xb

=0 0lim

=

limb

xb

e0

= + =

lim [ ]b

be 1 1

For n = 1:

u x dv e dx

du dx v e

x

x

= == =

xe dx xe dxx

b

xb

= lim 00=

+

limb

xb

xb xe e dx

0 0

=

+

lim limb b b

xbb

ee dx

0

=

+ =

limb be

11 1

For n = 2;u x dv e dx

du x dx v e

x

x

= == =

2

2

x e dx x e dxx

b

xb2 2

00

= lim

= +

limb

xb

xb x e x e dx2

0 02

=

+

lim limb b b

xbb

e xe d x

2

02

=

+

lim ( )b b

b

e

22 1

=

+ =

limb b

b

e

2 2 2

(b) Evaluate x e dxn x using integration by parts

u x dv e dx

du nx v e

n x

n x

= == =

1

x e dx x e nx e dxn x n x n x = + 1

f n x e dxn x( )+ =

1 0=

+

limb

n xb

n x x e nx e dx

0

1

0

=

+

limb

n

b

n xb

en x e dx1

0

= nf n( )

Note: apply L'Hpital's Rule times to shon ww that

lim .b

n

b

b

e

=

0


23/28

372 Chapter 8 Review

59. Continued

(c) Since f n nf n( ) ( ),+ =1

f n n n f n( ) ( ) ( ) !;+ = =1 1 1 thus

x e dxn x0 converges for all integers n 0.

60. (a) On a grapher, plot NINT sin , , ,xx

x x0 or create a table

of values. For large values ofx f x, ( ) appears to approach

approximately 1.57.

(b) Yes, the integral appears to converge.

61. (a)dx

x

dx

xb b1 12 2

11

+=

+ lim

=

=

lim tan

lim (tan tan )

b b

b

x

b

11

1 11

== + =

+=

+

4 2

3

4

1 121 21

dx

x

dx

xx

blim

=

=

=

lim tan

lim[tan tan ]

b

b

b

x

b

1

1

1 11

22 4 4

1

3

4 42

=

+= + =

dx

x

(b) f x dx f x dx f x dxc c

( ) ( ) ( ) = + 0

0

f x dx f x dx f x dxc c( ) ( ) ( )

= +0

0

Thus,

f x dx f x dxc

c( ) ( )

+

= + +

+ f x dx f x dx f x dx

f x d

c

c( ) ( ) ( )

( )

0 0

0

xx0

= +

f x dx f x dx( ) ( ) ,00

because

f x dx f x dx f x dx f x dxc

c c c

c( ) ( ) ( ) ( ) . + = =0 0

00

Quick Quiz Sections 8.3 and 8.4

1. E. lim.

lim.

lim.

. x x x

x

x

x

x

x

= = =

0 1 0 3

2

0 6

2

3

2

2

2. C. Sincedx

x pp1

1

1

= for p > 1, p > 0.

3. B.dx

xC

p+ =101

when p < 0.

4. (a)2

21

ln( )x

xdx

b

(b) limln

b

x

x

dx

2

21

(c) Note that2

2

lnx

xdx can be found by parts.

Let u x= 2ln and dv x dx= 2 .Then

22 2

2

1 1 1lnln ( ) ( )

x

xdx x x x x dx =

= +

= +

=

2 2

2 2

2

2

2

ln

ln

limln

x

x xdx

x

x xC

x

xbArea 11

1

2 2

2

b

b

b

b

dx

x

x x

b

b

=

=

limln

limln

+ +

=2

0 2 2b

Chapter 8 Review Exercises (pp. 470471)

1. 1 2 3 5 2 3 5 7, , ,

a40

401

40 1

40 341 43=

++

=( ) /

2. 3 6 12 24, , ,a40

393 2= ( )

3. (a) 1 2 1 3 2 =( ) /

(b) + =1 9 3 2 25 2( ) /

(c) a nn

n= + =

1 1 3 2

3 5

2( ) /

4. (a)

= 2

1 24

(b)1

24 20486( ) =

(c) ann n=

( ) ( )1 41

2

1

= ( ) ( )1 21 2 3n n


24/28

Chapte 8 Review 373

5.

[0, 20] by [2, 4]

6.

[0, 20] by [2, 2]

7. an

n

n

nn n n n=

+

= = =

lim lim lim ,3 1

2 1

6

4

3

2

3

2

2

2iit converges.

8. an

nn k

n

nn

n

n

n=

+=

+

>

( ) , , lim ( ) .13 1

22 1

3 1

20

an

nn k

n

nn

n

n

n=

+=

+


25/28


19. The limit leads to the indeterminate from 00. .

f

f

( ) (tan )

ln ( ) ln (tan )ln(tan )

/

=

= =1

limln(tan )

/ lim

sec

tan

lim

x x + +=

=

0 0

2

2

1 1

xx

x

+

+

=

+=

0

2

02 2

200

sin cos

limsin cos

lim (tan ) lim l n ( )

x x

fe e

+ += = =

0 0

0 1

20. lim sin limsin

limco

+ +

= =20

20

1

t t

t

t

ss t

t2=


x xx x

x xx

++

= +

=3 2

2

2

3 12 3

3 64 1

= 6 64

x


x x

x x

x

x x

+ +

=

=

3 1

2

6 1

4 3

2

4 3 3 2 =

6

12 60

2x x.

23. lim( )

( )lim

x x

f x

g x

x

x = =

5

1

5

fgrows at the same rate as g.

24. lim( )

( )lim

log

loglim

(ln )

x x x

f x

g x

x

x

x

= =2

3

/ /(ln )

(ln ) /(ln )

ln

ln

2

3

3

2x=


25. lim( )

( )lim lim

x x x

f x

g x

x

x x x =

+=

=

1

1

1 11

2


26. lim( )

( )lim

/lim

x x x x

xf x

g x

x

xe

e

= = =

100

100

fgrows faster than g.

27. lim( )

( )lim

tanx x

f x

g x

x

x = = 1 since

lim tan limx x

x x

= = 1

2

and


28. lim( )

( )lim

csc

/x x

f x

g x

x

x

=1

1

=

=

=

lim

lim

lim

x

x

x

x x

xx

x

x

x

1

1

1

1

1

2

2

2

2

2

==

=

lim/x x

1

1 11

2


29. lim( )

( )lim

ln

logx x

x

x

f x

g x

x

x =

2

=

==

lim

lim

lim

ln log

ln (ln ) /ln

x

x x

xx x

x

x

x

2

2

=

x

x

x

x

x

(ln )( / ln )

(ln )( /ln

lim

1 1 2

11

22 1

0

=)

Note that 11

2

ln< 0 since ln 2 < 1.

fgrows slower than g.

30. lim( )

( )lim lim lim

x x

x

x x

x

x x

f x

g x

= = =

3

2

2

3

2

3

x

=


26/28

Chapte 8 Review 375

33. lim( )

( )lim

tan ( )

x x

f x

g x

x

x

=1 1

1

=+

lim( )

( )

x

xx

x

1

1 1 22

2

=+

=

lim( )x x1

1 11

2


34. lim( )

( )lim

sin ( )

( )x x

f x

g x

x

x

=1

2

1

1

=

lim( )

( )

x

xx

x

1

1 1

2

2

2

3

=

=

lim( )x

x

x2 1 1 2


35. (a) lim ( ) limsin

x x

x

xf x

e =

0 0

2 1

1

= =

lim(ln ) (cos )

lnsin

x

x

x

x

e0

2 22

(b) Define f( ) ln .0 2=

36. (a) lim ( ) lim lnx x

f x x x + +

=0 0

= +lim

ln

x

x

x0 1

= +lim

x

xx0

211

= = +lim ( )

x

x0

0

(b) Define f( ) .0 0=

37.dx

x

dx

x xb

b

b

b

3 2 3 211 1

2/ /

lim lim= =

== 2

38.dx

x x21 7 12+ +

=+ +

= ++

lim

lim ln

b

b

b

dx

x x

x

x

21 7 12

4

3

=

1

5

4

b

ln

39.3

3 2

1 dx

x x

=

=

lim

lim ln

b b

b

dx

x x

x

x

3

3

3

2

1

=

b

1

2 2ln( )

40.dx

x

dx

x

x

b b9 9 32 3 20

3

3

1

0=

=

lim lim sinbb

0

3

2 =

41. ln( ) lim ln( ) lim ( ln ) x dx x dx x x xb b b

b

= = 0

1

0

1

0

1

= 1

42.dy

y

dy

y

dy

yb

b

b b2 3

02 31 0

2 3

1

1 / / /lim lim= + +

11

=

+

=

+lim lim

b

b

bb

y y01

0

1

1

2 26

43.d

( ) /+ 1 3 520

=+

++ +lim ( ) lim ( )/ /b

b

b b

d d

13 52 1

3 51 1

00

= +

+

+

lim ( ) lim (b

b

b

x x1

2

5

21

1

5

21

5

2++

=1 02

5

0

)

b

44.2

2

2

22 233

dx

x x

dx

x xb

b

=

lim

=

=

limb

bx

xIn In 3

2

3

45. x e dx x e dxx

b

xb2 2

00

= lim

= =

lim (( ) )

b

xb

x x e20

2 2 2

46. xe dxx30

= =

lim limb

x

b b

x xe dx

xe

30 3

3

1

9 =

b

01

9

47.dx

e e

dx

e e

dx

ex x b x xb b x+

=+

++

lim lim0

eex

b

0

=

lim (tan )b

xb

e1

0


27/28


47. Continued

+

l im (tan ( ))

b

xb

e1

0

=

2

48.4

16

4

16

42 2

0

2

dx

x

dx

x

dx

xb b b+=

++

lim lim ++ 160

b

=

lim tan

bb

x10

4

+

lim tanb

b

x1

04

=

49. 011

ln

lnz

zdz z dz

ln lim ln lim ( ln ) z dz z dz z z zb

b

b

b

= = =

1 11lnz

zdz

1

diverges

50. 011

e

tdt e dt

tt

e dt e dt e et

b

tb

b

tb

= = = +1 11

0lim lim ( ) == e 1

e

tdt

t

1 converges

51. (a)

=

3 8

3 1 2

1 3/

/

/

= 31

1 26

/

(b)1

2

(c) an

n

=

61

2

ann= 3 22( )

52. (a)5 5 11 5

4

1 5. .

.

=

11 5 1 5 13. ( . ) =

(b) 1 5.

(c) a nn = + 13 1 1 5( )( . )

an

= +1 5 14 5. .

53. (a) e dx e dx e dxx

b

x

b b

xb

= +2 20

2

0lim lim

(b) lim limb

x

b b

xb

e dx e dx

+20 2

0

=

+

=

lim lim

b

x

b b

xb

e e20

2

02 2

0++ + + =1

2

01

2

1

54. For x y 0 0, on (0,1].

Volume = ( l n ) x dx201

= (ln ) x dx201

= +

lim (ln )b

b x dx

0

21

Evaluate (ln ) x dx2 by integration by parts.u x= (ln )2 dv dx=

dux

x

dx=2(ln )

v x=

(ln ) (ln ) ln x dx x x x dx2 2

2 = Evaluate 2lnxdx using integration by parts.

dv dx=u x= 2ln

dux

dx=2

v x=

2 2 2 2 2ln ln ln x dx x x dx x x x C = = +(ln ) (ln ) ln x dx x x x x x C

2 22 2 = + +

Area = +

+ lim (ln ) ln

b b

x x x x x

0

21

2 2

= + +

lim (ln ) lnb

b b b b b0

22 2 2

= + + +

21

210

2

0

lim(ln )

limln

b b

b

b

b

b

=

+ + +

22 1

12

10 2 0 2

lim

(ln )( )lim

b b

b b

b

b

b

=

+ + +

22

12

0 0

lim(ln )

lim ( )b b

b

bb

= = + +

22

12 2

02

0

lim limb b

b

bb

55. For x y )0 0 0, , .on

Area0

= =

xe dx xe dxxb

xblim0

Evaluate xe dxx by using integration by parts.

u x dv e dx

du dx v e

x

x

= == =


28/28

Section 9.1 377

55. Continued

xe dx xe e dx xe e C x x x x x = + = +

Area =

= +

lim

lim

b

x xb

b

b b

xe e

be e

0

11

1

11 1

= +

= + =

lim

lim

b b

b b

be

e

56. (a) xe dxx 20

(b) limb

xb xe dx

20

(c) Note that xe dxx 2 can be found by parts: xe dx x e e dx x x x = 2 2 22 2( ) ( )

= + 2 42 2 xe e C x x .

Area =

lim limb

xb

b

x xb

xe dx xe e

= 2

0

2 2

02 4

= +( ) =

lim .b

b bbe e2 4 0 4 42 2

57. (a) x dydy

y

2

200 1=

+

( )

(b) lim( )b

b dy

y + 1 20

(c) Volume

= + = +

lim

( )lim ( )

b b

bb dy

yy

11

2

1

00

= +

+

=

limb b

1

11

58. Note that xe dxx can be found by parts:

xe dx x e e dx xe e C x x x x x = = + ( ) ( ) .So

xe dx xe dx xe ex

k

x

k

x xkk

= = lim lim 00

0

= +

limk k k

k

e e

11

By LHopitals rule, lim lim .k k k k

k

e e

=

=1 0

Therefore, xe dxk

e e

x

k k k

= +

lim

11

0

= + =0 0 1 1The integral converges to 1.

Chapter 9

Infinite Series

Section 9.1Power Series (pp. 473 483)

Exploration 1 Finding Power Series for

Other Functions1. 1 2 3 + + + + x x x x n ( ) .

2. x x x x xn n + + + ++2 3 4 11 ( ) .

3. 1 2 4 8 22 3+ + + + + + x x x x n ( ) .

4. 1 1 1 1 1 12 3 + + + +( ) ( ) ( ) ( ) ( ) . x x x xn n This geometric series converges for 1

ch. 8 solutions new calc book

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