350 Section 8.1

(1)

51. y′ =sec2x, so the area is 2 1 2 2

0 4

π

tanx sec x dx,

(

)

+

( )

∫

which using NINT evaluates to ≈3 84. .

52.x= 1 1 2 1 1 1

2 2

y andx′= −y so the area is y y ⎛

⎝⎜ ⎞

⎠⎟ + −

⎛ ⎝⎜

, π ⎞⎞

⎠⎟

∫

2

1 2

dy,

which using NINT evaluates to ≈ 5.02.

53. (a) The two curves intersect at x= 1.2237831. Store this value as A.

Area = (2 sin sec ) 1 366. .

0 + − =

∫

A x x dx

(b)Volume = π (2 sin )2 (sec )2 16 404. .

0

(

+ −

)

=

∫

A x x dx

(c)Volume = (2 sin sec )2 1 629. .

0 + − =

∫

A x x dx

54. (a) Average temp

=

− −

⎛ ⎝⎜ ⎞⎠⎟ ⎛

⎝⎜

⎞

⎠⎟ ≈ °

∫

1

14 6 6 80 10 12 87

14

cos πt dt F.

(b) F t( )= − cos⎛ t

⎝⎜ ⎞ ⎠⎟≥

80 10

12 78

π _for

5 2308694. ≤ ≤t 18 766913. . Store these two values as A and B.

(c)Cost = − ⎛

⎝⎜ ⎞ ⎠⎟− ⎛

⎝⎜

⎞ ⎠⎟ ≈

∫

0 05 80 10

12 78 5 10

. cos πt dt .

A B

The cost was about $5.10.

55. (a) 15600

24 160 6004

2 9 17

(t − t+ )dt≈

∫

people.

(b)

15 15600

24 160

11 15600

24 16

2 9

17

2

( )

(

t t dt

t t

− +

+

− +

∫

0

0 104 048

17 23

)dt≈ ,

∫

dollars

(c) H′(17)=E(17)−L(17)≈ −380 people. H(17) is the number of people in the park at 5:00, and H′(17 is the) rate 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, L’Hôpital’s Rule,

and Improper Integrals

Section 8.1

Sequences (pp. 435–443)

Quick Review 8.1

1. f( )5 5 5 3

5 8

= =

+

2. f(− = −)

− + = −

2 2

2 3 2

3. − + −2 (3 1 1 5)( . )=1 4. − + −7 (5 1 3)( )=5 5. 1 5 2. ( 4 1−)=12 6. −2 1 5( . 3 1−)= −4 5.

7. lim lim

x x

→∞ →∞

+

+ = =

5 2

3 16

5

3 0

3 2

4 2

3 4

8. limsin ( ) lim

x x

→0 = →0 =

3 3

3

9. lim sin lim

x→∞ x x x→∞ x x ⎛

⎝⎜ ⎞⎠⎟ ⎛

⎝⎜

⎞ ⎠⎟=

⎛ ⎝⎜ ⎞⎠⎟=

1 1

1

10. lim .

x

x x x

x x

→∞

+

+ = = ∞

2 1

2

3 2 3

Does not exist, or

Section 8.1 Exercises

1. 1 2

2 3

3 4

4 5

5 6

6 7

50 51

, , , , , ;

2. 2 5 2

8 3

11 4

14 5

17 6

149 50

, , , , , ;

3. 2 9 4

64 27

625 256

7776

3125 2 48832

, , , , ≈ . ,

117649

46656 2 521626 51

50 2 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+7d= − +2 7 3( )=19 (c) a_n=a_n₋₁+3

(d) a_n= − + −2 (n 1 3)( )=3n−5 12. (a) −2

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

(d) a_n= + − − = − +15 (n 1)( 2) 2n 17 13. (a) 1

2

(b) a+ d= + ⎛ ⎝⎜

⎞ ⎠⎟=

7 1 7 1

(2)

13. Continued (c) a_n=a_n₋₁+1

2

(d) a_n= + − ⎛n n

⎝⎜ ⎞⎠⎟=

1 1 1

2 1 2

( ) ( + )

14. (a) 0.1

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

(d) a_n= + −3 (n 1 0 1)( . )=0 1. n+2 9. 15. (a) 1

2

(b) 8 1

2 0 03125

8 ⎛ ⎝⎜

⎞ ⎠⎟ = .

(c) a_n= ⎛ a_n ⎝⎜

⎞ ⎠⎟ − 1

2 1

(d) a_n

n n

= ⎛

⎝⎜ ⎞⎠⎟ = −

8 1

2 2

1 4

−

16. (a) 1 5.

(b) ( )( . )1 1 58≈25 6289. (c) a_n=( . )1 5 a_n₋₁

(d) a_n=( )( . )1 1 5n−1=( . )1 5n−1 17. (a) −3

(b) (−3)9= −19 683, (c) a_n= −3a_n₋₁

(d) a_n= − −( 3)( 3)n−1= −( 3)n 18. (a) −1

(b) ( )(5 − =1)8 5 (c) a_n= −a_n₋₁

(d) a_n= −5( 1)n−1 19. 7 2

3 3

− − =( )

a₁= − − = −2 3 5

a_n=a_n₋₁+3for all n≥2

20. − − =3 5

4 −2

a₁= − −5 ( 2 4)( )=13

a_n= + − − = − +13 (n 1)( 2) 2n 15

21.r= ⎛

⎝⎜ ⎞⎠⎟ =

3010000

3010 10

1 3/

a₁ 3

3010

10 3 01

= = .

a_n=3 01 10. ( )n−1,n≥1

22.r= − ⎛ ⎝⎜

⎞ ⎠⎟ =

16

1 2 2

1 5

/

/ −

a₁ 1 2

2 1 4

=−

− =

/ /

a_n= −( 1)n−1( )2n−3,n≥1 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.

(3)

29.

[0, 20] by [–1, 5]

30.

[0, 10] by [–15, 10]

31. lim lim ( ) lim

n n n

n

n n

→∞ + = →∞ + →∞

⎛ ⎝⎜

⎞

⎠⎟= =

3 1

3 1 3 0− 3

converges, 3

32. lim lim

n n

→∞ + = →∞ = 2

3 2

2 converges, 2

33. lim lim

n n

n n n n

n n

→∞ →∞

−

+ + = =

2 1

5 2

2 5

2 2

2 2 −

converges, 2 5/

34. lim lim lim

n n n

n n

n

n n

→∞ 2₊ = →∞ 2= →∞ = 1

1 0 converges, 0

35.n k n

n

n n

= − −

+ =

→∞

2 1 1

3 1

, lim ( )

n k n

n

n n

= − − −

+ = −

→∞

2 1 1 1

3 1

, lim ( ) diverges

36.n k n

n

n n

= − +

+ = − =

→∞ →∞

2 1 1

1 1 0

2 2

, lim ( ) lim ( )

n k n

n

n n

= − − +

+ = − =

→∞ →∞

2 1 1 1

1 1

2 2

, lim ( ) lim ( ) 00

converges, 0 37. lim ( . )

n n

→∞1 1 = ∞ diverges 38. lim ( . )

n

→∞ 0 9 =0 converges, 0

39. lim sin lim

n→∞ n n n→∞ nn ⎛

⎝⎜ ⎞ ⎠⎟ ⎛

⎝⎜

⎞ ⎠⎟=

⎛ ⎝⎜

⎞ ⎠⎟=

1 1

1 converges, 1

40. lim cos

n

→∞

⎛ ⎝⎜ ⎞⎠⎟ ⎛

⎝⎜

⎞ ⎠⎟ π

2

− ≤ ⎛

⎝⎜ ⎞ ⎠⎟≤

1

2 1

cos nπ ,diverges

41. lim sin

n

n n

→∞

⎛ ⎝⎜ ⎞⎠⎟

− ≤1 ≤1

n n

sin

lim lim

n→∞ −n n→∞ n ⎛

⎝⎜ ⎞⎠⎟= ⎛⎝⎜ ⎞⎠⎟=

1 1

0

42. lim

n→∞ n

⎛ ⎝⎜ ⎞⎠⎟

1 2

lim lim

n→∞ −n n→∞ n ⎛

⎝⎜ ⎞ ⎠⎟=

⎛ ⎝⎜

⎞ ⎠⎟=

1 1

0

Note: 1

2n < nforn≥ ⎛

⎝⎜ ⎞⎠⎟

1

1 .

43. lim !

n→∞ n ⎛ ⎝⎜ ⎞⎠⎟

1

lim lim

n→∞ −n n→∞ n ⎛

⎝⎜ ⎞ ⎠⎟=

⎛ ⎝⎜

⎞ ⎠⎟=

1 1

0

Note:1

! for

n ≤n n≥ ⎛

⎝⎜ ⎞⎠⎟

1

44. lim sin

n n

n

→∞

⎛ ⎝ ⎜ ⎞_⎠⎟

2

lim lim

n→∞ −n n→∞ n ⎛

⎝⎜ ⎞⎠⎟= ⎛⎝⎜ ⎞⎠⎟=

1 1

0

Note: 1

2n <nforn≥1 ⎛

⎝⎜ ⎞⎠⎟

1

45.Graph (b) 46.Graph (c) 47.Table (d) 48.Table (a)

49.False. Consider the sequence withnth term

a_n=− +5 2(n−1). Here

a= −5,a₂= −3,a₃= −1,and a₄=1.

50.True. a a

a

r 1

1

0 2 0

> , = > , and

a_n=a r₁ n−1>0 for all n≥2.

51.C. 5− −( 1)=

2 3

− +1 3 5( )=14

52.E. 1 25 2 5

1 2 .

. = 2 5

1 2 5

. / = 53.D. lim sin

n→∞ n n

⎛ ⎝⎜ ⎞⎠⎟ ⎛

⎝⎜

⎞ ⎠⎟

3π

= ⎛

⎝⎜ ⎞ ⎠⎟=

→∞ lim

n n n

3 3

(4)

54.E. n k n n

n

= − −

+ ⎛

⎝⎜

⎞ ⎠⎟=

→∞

2 1 3 1

2 1

, lim ( )

n k n

n

= − − −

+ ⎛

⎝⎜

⎞ ⎠⎟= −

→∞

2 1 1 3 1

2 1

, lim ( )

55. (b) lim sin

n→∞ n n

⎛ ⎝⎜ ⎞⎠⎟ ⎛

⎝⎜

⎞ ⎠⎟

2 π

= ⎛

⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜

⎞ ⎠⎟=

→∞ lim

n

n n

2 π 2π

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

[0, 10] by [–10, 60]

57. a_n =arn−1implies that loga_n=loga+ −(n 1) log .r Thus loga_n

{

}

is an arithmetic sequence with first term log a and common ratio log r.

58. a_n= + −a (n 1)dimplies that

10an =10a+ −(n 1)d=10 10a( d n) −1. Thus 10

{ }

an is a geometric sequence with first term10aand common ratio10d. 59. Givenε >0 chooseM=1

ε. Then

1 0

n− <εifn M> .

Section 8.2

L’Hôpital’s Rule (pp. 444–452)

Exploration 1 Exploring L’Hôpital’s Rule Graphically

1. limsin limcos

x x

x

→0 = →0 1 =1

2. The two graphs suggest that lim lim .

x x

y y

→ = →

′ ′ 0

1

2 0

1 2

3. y x x x x

5= 2

cos sin .

−

The graphs ofy₃andy₅clearly show

that l’Hôpital’s Rule does not say that lim

x

y y

→0 1 2

is equal to

lim .

x

y y

→

⎛ ⎝⎜

⎞ ⎠⎟ ′

0 1 2

Quick Review 8.2 1.

x ⎛1+0 1

⎝⎜ ⎞⎠⎟

.

x

1 1.1000

10 1.1046

100 1.1051

1000 1.1052

10,000 1.1052

1,000,000 1.1052

Asx

x

→ ∞ ⎛ + ⎝⎜ ⎞⎠⎟

, 1 0 1. approach 1.1052.

2. _x _x1/(ln )x

0.1 2.7183

0.01 2.7183

0.001 2.7183

0.0001 2.7183

0.00001 2.7183

Asx→0+,x1/(ln )x approaches 2.7183. 3.

x ⎛1−1

⎝⎜ 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

Asx

x

→ ⎛ −

⎝⎜ ⎞ ⎠⎟

−

0 , 1 1 approaches 1. 4.

x ⎛1+1

⎝⎜ ⎞ ⎠⎟ x

x

–1.1 13.981

–1.01 105.77

–1.001 1007.9

–1.0001 10010

Asx

x

→ ⎛ +

⎝⎜ ⎞⎠⎟

−

−1 , 1 1 goes to∞.

5.

Ast t

t

→ −

−

1 1

1

(5)

6.

Asx x approaches 2.

x

→ ∞ +

+

, 4 1

1

2

7.

Asx xapproaches .

x

→0,sin3 3

8.

Asθ π θ approaches1.

θ →

+

2 2, tan

tan 9. y

h h = 1sin

10. y= +(1 h)1/h

1. lim

x

x x

→ ₋

⎛ ⎝⎜

⎞ ⎠⎟

2 2

2 4

− _{appears to be about}1 4;

[–2, 4] by [–1, 4]

By L’Hôpital’s Rule:

lim lim

( )

x x

→ = →

− − ⎛ ⎝⎜

⎞

⎠⎟= =

2 2 2

2 4

1 2 2

1 4

2. lim sin ( )

x

x x

→ = ⎛⎝⎜

⎞ ⎠⎟ 0

5

appears to be about 5;

[–2, 2] by [–2, 6]

lim sin ( ) lim cos( ( ))

x x

→ →

⎛ ⎝⎜

⎞ ⎠⎟=

⎛ ⎝⎜

⎞ ⎠

0 0

5 5 5 0

1 ⎟⎟ =5

3. lim

x

x x

→

+ − − ⎛ ⎝

⎜ ⎞

⎠ ⎟ 2

2 2

2 appears to be about 1 4;

[–2, 4] by [–1, 1]

lim lim

( )

x x

→ →

−

+ − − ⎛ ⎝

⎜ ⎞

⎠

⎟ = +

⎛

⎝ ⎜ ⎜ ⎜

2 2

1 2

2 2

2

1 2 2 2

1

⎞⎞

⎠ ⎟ ⎟ ⎟=

1 4

4. lim

x

x x

→

− − ⎛ ⎝

⎜ ⎞

⎠ ⎟ 1

3 ₁

1 appears to be about 1 3;

[–2, 2] by [–1, 2]

lim lim

( )

x x

→ →

−

− − ⎛ ⎝

⎜ ⎞

⎠ ⎟ =

⎛

⎝ ⎜ ⎜ ⎜

⎞

⎠ ⎟ 1

3

1 2 3

1 2

1 31

1 ⎟⎟_⎟= 1 3

5. lim cos lim sin lim

x x

→ →

− ⎛

⎝⎜ ⎞⎠⎟=

⎛ ⎝⎜

⎞ ⎠⎟=

0 2 0

1

2 xx→

⎛

⎝⎜ ⎞⎠⎟= 0

0 2

1 2 cos( )

6. lim

sin

cos( ) lim

cos

/ /

θ π θ π

θ θ

θ

→ →

− + ⎛ ⎝⎜

⎞ ⎠⎟=

−

2 2

1

1 2 −−

⎛ ⎝⎜

⎞ ⎠⎟

= −

− ⎛

⎝⎜ ⎞

→

2 2

2

4 2

2

sin( )

lim

sin

cos

/

θ π

π

θ π

⎠⎠⎟ ⎛

⎝ ⎜ ⎜ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ⎟ ⎟

=1

4

7. lim cos lim sin

t t t t

t e t

t e

→ →

− − − ⎛ ⎝⎜

⎞ ⎠⎟=

− − ⎛ ⎝⎜

0 0

1

1 1

⎞⎞ ⎠⎟=

− ⎛

⎝⎜ ⎞⎠⎟= −

→

lim cos

t 0 e0

0 1

8. lim lim

x x

→ →

− +

⎛ ⎝

⎜ ⎞_⎠⎟ = −₋

2 2

3 ₂ 2

4 4

12 16

2 4

3 112

2 6 2

1 6

2 ⎛

⎝⎜

⎞ ⎠⎟=

⎛ ⎝⎜

⎞ ⎠⎟=

→ lim

( )

x

9. (a) lim sin

sin lim

cos( ( )) c

x x

→− →−

⎛ ⎝⎜

⎞ ⎠⎟=

0 0

4 2

4 4 0

2 oos( ( ))2 0 2

⎛ ⎝⎜

⎞ ⎠⎟=

(b) lim sin

sin lim

cos( ( )) c

x x

→+ →+

⎛ ⎝⎜

⎞ ⎠⎟=

0 0

4 2

4 4 0

2 oos( ( ))2 0 2

⎛ ⎝⎜

(6)

9. Continued (b)

[–2, 2] by [–3, 3]

10. (a) lim tan lim sec

x x

x

→− → −

⎛ ⎝⎜

⎞ ⎠⎟=

⎛ ⎝

⎜ ⎞_⎠⎟ =

0 0

2

1 1

(b) lim tan lim sec

x x

x

→+ →+

⎛ ⎝⎜

⎞ ⎠⎟=

⎛ ⎝

⎜ ⎞_⎠⎟ =

0 0

2

1 1

[–1.5, 1.5] by [–2, 10]

11. (a) lim sin lim cos( ) ( )

x→− x x→ −

⎛ ⎝⎜

⎞ ⎠⎟=

⎛ ⎝⎜

⎞ ⎠

0 3 0 2

0 3 0

x

⎟⎟ = ∞

(b) lim sin lim cos( ) ( )

x x

→+ →+

⎛ ⎝⎜ ⎞⎠⎟=

⎛ ⎝⎜

⎞ ⎠

0 3 0 2

0 3 0 ⎟⎟ = ∞

[–2, 2] by [–2, 10]

12. (a) lim tan lim sec ( ) ( )

x x

→− → −

⎛

⎝⎜ ⎞⎠⎟= −

⎛ ⎝

⎜ ⎞

0 2 0

2 ₀

2 0 _⎠⎠⎟ = −∞

(b) lim tan lim sec ( ) ( )

x x

→+ → +

⎛ ⎝⎜

⎞

⎠⎟= +

⎛ ⎝

⎜ ⎞

0 2 0

2

0 2 0 _⎠⎠⎟ = ∞

[–1, 1] by [–10, 10]

13.Left:

lim csc

cot

x

x x

→ − +

⎛ ⎝⎜

⎞ ⎠⎟= ∞−∞

π 1 lim csc cot

csc

x

x x x

→ −

− − ⎛ ⎝⎜

⎞ ⎠⎟= −

π 2 1

Right:

lim csc

cot

x

x x

→ + +

⎛ ⎝⎜

⎞ ⎠⎟= −∞∞

π 1 lim csc cot

csc

x

x x x

→ +

− − ⎛ ⎝⎜

⎞ ⎠⎟= −

π 2 1

limit= −1

[3π/4, 5π/4] by [–5, 5]

14.Left:

lim sec

tan

/

x

x x

→ −

+ ⎛ ⎝⎜

⎞ ⎠⎟= ∞∞

π2

1

lim sec tan sec

/

x

x x x

→ −

⎛ ⎝⎜

⎞ ⎠⎟=

π2 2

1 Right:

lim sec

tan

/

x

x x

→ +

+ ⎛ ⎝⎜

⎞ ⎠⎟= −∞−∞

π2

1

lim sec tan sec

/

x

x x x

→ +

⎛ ⎝⎜

⎞ ⎠⎟=

π2 2

1 limit= 1

[π/4, 3π/4] by [–5, 5]

15. limln ( ) log

x

x x

→∞ + = ∞∞

1

2

lim ln

ln

x

→∞

+ ⎛

⎝ ⎜ ⎜ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ⎟ ⎟ =

1 1 1

2 2

[0, 100] by [–1, 2]

16. lim

x

x x

x

→∞

− + ⎛ ⎝

⎜5₇ 3₁⎞_⎠⎟ =∞_∞ 2

2

lim

x

x x

→∞

− ⎛

⎝⎜ ⎞⎠⎟=

10 3

14 5 7

[0, 100] by [–1, 2]

17. lim ( ln )

x

x x

→0+ = ∞

0i

lim ln

/ lim

/

/ l

x x

→ + →+

⎛ ⎝⎜

⎞

⎠⎟= −

⎛ ⎝⎜

⎞ ⎠⎟=

0 1 0 2

1

1 xiim ( )

x

→0+ − =

(7)

18. lim tan

x→∞ x x ⎛ ⎝⎜ ⎞⎠⎟ ⎛

⎝⎜

⎞ ⎠⎟= ∞

1

0 i

lim tan ( ) lim

x h h

h

hh

→∞ →∞

⎛ ⎝⎜

⎞ ⎠⎟=

⎛ ⎝⎜

⎞ ⎠⎟=

1 1

1

19. lim (csc cot cos )

x

x x x

→0+ − + = ∞− ∞

lim cos cos sin

sin lim

sin

x x

x x x

x

→ + → +

− +

⎛ ⎝⎜

⎞ ⎠⎟=

0 0

1 xx x x

x

− +

⎛ ⎝

⎜ sin_cos cos ⎞_⎠⎟ =

2 2

1

20. lim (ln( ) ln( ))

x→∞ 2x − x+1 = ∞− ∞ lim ln( )

ln( ) ln( )

x

x x

→∞ +

⎛ ⎝⎜

⎞ ⎠⎟=

2

1 2

21. lim ( )/ ( )

x

x x

e x

→

∞ ∞

+ = + =

0

1 _{1 0} ₁

lim ln( ) lim ln li

x

e x x

x x

x

→ →

+ ⎛ ⎝

⎜ ⎞_⎠⎟ = ⎛_⎝⎜ + ⎞_⎠⎟=

0 0 xmm

x

→

+ ⎛

⎝ ⎜ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ⎟= 0

1 1

1 2

lim ln ( )

x f x

e e

→0 = 2

22.lim ( ( ))

x x

x

→

− ₌ ∞

1

1 1 ₁

limln lim /

x x

x

→1 −1= →1 =

1

1 1

lim ln ( )

x f x

e e e

→1 = = 1

23.lim ( ) ( ( ) )

x

x x

→

− −

− + = − + =

1

2 ₂ ₁ 1 ₁2 _{2 1} ₁1 1 ₀0

limln ( )

/ lim /( )

x x

x

x x

→ →

− +

− =

−

− − =

1 2

2 1

1 1

2 1

1 1 llim

( )

x

x x

→

− −

− ⎛ ⎝

⎜ ⎞_⎠⎟ =

1

2

2 1

1 0

lim ln ( )

x f x

e e

→1 = =

0 ₁

24. lim (sin )

x

→0+ = 0

0

lim ln (sin )

/ lim tan

/

x x

x

x x

x

→ + = →+

− ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟

0 1 0 ₂

1

1 ⎟⎟

= ⎛−

⎝ ⎜ ⎞_⎠⎟

= ⎛−

⎝⎜

⎞

→

+

lim tan

lim sec

x

x x

0 2

2

⎠⎠⎟= − =

0

1 0

lim ln ( )

x

f x

e e

→ +0 = = 0

1

25. lim ( )

x

→ + +

⎛ ⎝⎜

⎞

⎠⎟ = + ∞ = ∞

0

0 0

1 1 1

lim lim

ln ( )

/ lim

x

x x

x

x x

→ + + → + →

⎛ ⎝⎜

⎞ ⎠⎟ =

+ =

0 0

1 1

1 00 2

0 2

1 1 1

1

+

→ +

− + − ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

=

+

x x x x

x x

x

( )

/ lim

( )

=

+ =

→ + lim

x

x x

0 1

0

lim ln ( )

x

f x

e e

→0+ = = 0

1

26. lim (ln )/

x

→∞

∞

= ∞ 1

limln lim /

x x

x

→∞ = →∞ =

1

1 0

lim ln ( )

x f x

e 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 lim ln lim /

x x x

x x

x

→∞ = →∞ = →∞ = =

5 ₅ ₅

1 0

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) lim sin lim cos

x x

x

x x

→ + → +

− ₌ −

0 3 0 2

1 3

=

→

+

lim sin lim cos

x

x x

x 0

0

6

6 1 6 29.Letf( ) sin

sin .

θ θ

θ

= 3

4

θ ±100 ±10−1 ±10−2 ±10−3 ±10−4

f( )θ −0.1865 0.7589 0.7501 0.7500 0.7500

Estimate the limit to be =3 4. lim

sin lim

cos cos

θ θ

→0 4 = →0 =

3 3

4 4

3 4 sin 3

30. Letf t

t

t t t t

( ) sin

sin

= 1 − = −1

sint .

t ±100 ±10−1 ±10−2 ±10−3

f(t) ±0.1884 ±0.0167 ±0.0017 ±0.00017

Estimate the limit to be 0. lim

sin

sin sin lim

t t

t

t t

t t t t

→ − →

⎛ ⎝⎜

⎞ ⎠⎟=

−

=

0 0

1 1

lim

→ →

→

− + =

− + +

0

1 cos cos sin

lim sin

sin cos c

t

t t t

t

t t t

(8)

31. Letf x( )= +(1 x)1/x.

x 10 102 103 104 105

f(x) 1.2710 1.0472 1.0069 1.0009 1.0001

Estimate the limit to be 1. ln ( ) ln ( )

limln ( ) lim

f x x

x

x x

x

x x

= +

+ ₌ _{+ =}

→∞ →∞

1

1 1

1 0 1

1 0

1 1

=

+ = =

→∞ →∞ →∞

lim ( )/ lim ( ) lim ln ( )

x

x x

f x

x f x e ==e0=1

32.Letf x x 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

→∞ →∞ →∞

−

+ =

−

+ = − =

2

3 5

1 4

6 5

4 6

2

2 −−

2 3.

33. limsin lim cos ( )( ) cos( )

θ θ

→0 = → = =

2 0

2

1 2 0 0 0

34. lim

ln sin lim

cos

t t

t

t t

→ →

−

− = ₋

1 1

1

π _π _π

= − −

1 1 π( 1)

= +

1 1

π

35. lim log

log ( ) lim

ln

( ) ln

x x

x

→∞ + = →∞

+ 2

3 3

1 2 1

3 3

= +

→∞ lim( ) ln

ln

x

x x

3 3

2

= +

→∞

lim ln ln

ln

x

x x

3 3 3

2

=

→∞ limln

ln

x

3 2

= ln

ln 3 2

36. lim ln ( )

ln lim

y y

y

y y

y

→ + → +

+ ₌

+ + 0

2

0 2

2

2 2

2 1

= +

+

= +

+

→

+

lim ( )

y

y y

0 2

0 2 2

2 2

2

2 2

2

= +

+

→ + lim

y

y y 0

4 2

2 2

= +

+ = =

4 0 2

2 0 2

2

2 1

( ) ( )

37. lim tan lim

si

/ /

y y y y

y

→ − →

⎛

⎝⎜ ⎞⎠⎟ =

⎛ ⎝⎜ ⎞⎠⎟

π π

2 2 2

2− nn cos

y

=

− ⎛

⎝⎜ ⎞⎠⎟ + −

−

→lim

cos ( )sin sin

/

y

y y y

y

π

2

2 1

= − ⎛

⎝⎜ ⎞⎠⎟ + −

−

π π π π

π

2 2 2 1 2

2 cos ( )sin

sin

= −

− =

( )( ) ( ) 1 1

1 1

38. lim ( sin ) lim

sin

x x

x x x

x

→0+ − = →0+

1n 1n 1n

Letf x x

x

( ) sin .

=

lim

sin lim

lim

x x

x

x x

→ →

→

+ = + =

0 0

1 1

cos Therefore,

0

0 0

1 1 0

+(lnx− nsin )x =_xlim ln ( )_→ + f x =ln =

39. lim lim

x x x x

x x

→ + − →+

⎛ ⎝⎜

⎞ ⎠⎟=

− _{= ∞}

0 0

1 1 1

40.The limit leads to the indeterminate form∞0. Letf x

x

( )= ⎛ .

⎝⎜ ⎞ ⎠⎟

1

2

ln ln

ln

1 1

1

2 2

2

x x x

x

⎛

⎝⎜ ⎞⎠⎟ = ⎛⎝⎜ ⎞⎠⎟= ⎛ ⎝⎜ ⎞⎠⎟

lim ln

lim / /

/ li

x x

x

x x

x

→ →

⎛

⎝⎜ ⎞⎠⎟₌ −

− =

0 2

0 3 2

2

1

2 1

1 x→mm02x=0

lim lim ln ( )

x

x f x

x e e

→ →

⎛

⎝⎜ ⎞⎠⎟ = = =

0 2 0

0

1

41. lim lim

x x

x

x x x

→±∞ →±∞

−

− + = − =

3 5

2 2

3

4 1 0

(9)

42.lim sin

tan lim

cos sec

x x

→0 = →0 2 =

7 11

7 7

11 11

7 11

43.The limit leads to the indeterminate form∞0. Letf x( )= +(1 2x)1 2/( ln )x.

ln ( ) ( )

ln

/( ln )

1 2 1 2

2

1 2

+ x = + x

x

x ln

limln ( )

ln lim lim

x x x

x x

x

→∞ →∞ →∞

+ ₌ + ₌

+

1 2 2

2 1 2

2 1 2xx=xlim→∞ = 1 2

1 2

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

x

x x

f x

x e e e

→∞1 2+ = →∞ = =

1 2 1 2

44. The limit leads to the indeterminate form00. Letf x( )=(cos )x cosx.

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

cos

x x x x

x

x₌ ₌

lim ln (cos )

sec lim

sin cos sec

/ /

x x

→ − = → −

−

π2 π2 xxtanx

= −

→lim−

tan sec tan

/

x

x x x

π2

= − =

→lim_/− cos

x

π2

0

lim (cos )cos lim ln ( )

x

x x

f x

x e e

→ − = −

= = = =

π π

2 2

0

1

45.The limit leads to the indeterminate form1∞. Letf x( )= +(1 x)1/x.

ln 1+ ln 1+

1+

( ) ( )

lim ln ( ) lim 1

/

x x

x

x x

1

0 0

=

→ + →+

x

1 1

0 1

0

1

+ ₌

= = =

→ + →+

x

x e e e

x

x x

f x

lim (1+ )/ lim ln ( )

46. The limit leads to the indeterminate form Letf x( )=(sin )x tanx

ln sin ln sin ln sin

cot

( ) tan ( ) ( )

lim

tan

x x x x

x

= =

→00+ = →0+− 2 = → ln sin

cot

( )

lim cos sin csc

lim

x x

x x x

x x 00

0 0

0

+

+ +

− =

=

→ →

( sin cos ) lim (sin )tan lim l

x x

x e

x

x x

n n ( )f x

e = 0=

1

47. The limit leads to the indeterminate form1−∞. Letf x( )=x1 1/(−x).

lnx/( ) lnx x

x

1 1

1 − ₌

−

lim ln lim

x x

x

→1+1− = →1+− = −

1

1 1

lim /( ) lim ln ( )

x

x x

f x

x e e

e

→ −

→

−

+ = + = =

1 1 1

1

1 1

48. dt

t t x x

x x

x x x

x

= ⎡⎣ ⎤⎦ = − =

∫

2 ln 2 ln2 ln ln2

lim lim ln lim ln ln

x x x

x

dt t

x x

→∞

∫

= →∞ = →∞ =

2 2

49. lim lim /

x x

x x x

x x

→ →

−

− − = − =

1 3

3 ₁

2 2

1

4 3

3

12 1 3 11

50. lim lim lim

x x x

x x

→∞ →∞ →∞

+

+ + =

+

+ =

2 3

1

4 3

3 1

4 6

2

3 2 _xx=0

51. lim cos

limsin sin lim

x x

x

t dt

x

x x

→ →

∫

− =

−

− =

1 1

2 ₁ 2

1

1 xx

x x

→1 2 = 1 2

cos cos

52. lim limln ln lim /

x x

dt t x

x x

→ → →

∫

− =

−

− =

1 1

3 ₁ ₁ 3 ₁

1 1

1 xx x

3 2=1 3/

53. (a) L’Hôpital’s Rule does not help because applying L’Hôpital’s 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 L’Hôpital’s Rule a second time returns to the original problem.

lim lim( / )( )

( / )(

/

x x

→∞ →∞

−

+

+ =

+ +

9 1

1

9 2 9 1

1 2

1 1

9 1

1 2

)−/ = lim→∞

+ +

x

x x

(b)

The limit appears to be 3.

(c) lim lim

x x

x

→∞ →∞

+

+ =

+

= =

9 1

1

9 1

1 1 9

1 3

54. (a) L’Hôpital’s Rule does not help because applying L’Hôpital’s 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 L’Hôpital’s Rule a second time returns to the original problem.

lim sec

tan lim

sec tan

sec lim

/ /

x x x

x x

x x x

→π2 = →π2 2 = →→π/

tan sec

(10)

54. Continued (b)

The limit appears to be 1.

(c) lim sec

tan lim

cos sin cos

lim

/ /

x x x

x x

x x x

→π2 = →π2 =

1

→

→π/2sin = 1

1

x

55. Find c such that lim ( ) .

x

f x c

→0 =

lim ( ) lim sin

x f x x

x x

x

→ = →

−

0 0 3

9 3 3

5

= −

→

lim cos

x

x x

0 2

9 9 3

15

=

→ lim sin

x

x x 0

27 3

30

= = =

→ lim cos

x

x 0

81 3

30 81 30

27 10 Thus,c=27

10. This works since lim ( )x→0 f x = =c f( ),0 sof is

continuous. 56. f x( ) is defined atx

x

≠

→ 0

0

. lim f x( ) leads to the indeterminate form00.

ln ln ln

ln

x x x x

x

x x

= =

=

− =

→ →

1

1 1

0 0

2

lim lim lim

xx

x x

x

x e e

→

→ →

− =

= = =

0

0 0

0

1

lim lim ln x

Thus,f has a removable discontinuity at x= 0. Extend the definition ofƒ by lettingƒ(0)= 1.

57. (a) The limit leads to the indeterminate form1∞. Letƒ(k)=⎛1+

⎝⎜ ⎞⎠⎟ r k

kt

.

ln ln 1

ln 1 1

f k kt r k

t r

k

( )= ⎛

⎝⎜ ⎞ ⎠⎟=

+ ⎛ ⎝⎜ ⎞⎠⎟

lim lim

k k

t r

k

t r k

r k

→∞ →∞

+ ⎛

⎝⎜ ⎞⎠⎟ ₌ ⎛⎝⎜− ⎞⎠⎟ +

ln 1 1

1

2 ⎛⎛ ⎝⎜ ⎞⎠⎟ −

−1

2

1

k

=

+ = =

→∞ lim

k

r t r k

r t rt

1 1

lim lim

k

k t

k

k t

A r

k A

r k

→∞ + →∞

⎛ ⎝⎜

⎞

⎠⎟ = +

⎛ ⎝⎜

⎞ ⎠⎟

0 1 0 1

=

→∞

A e

k f k

0lim ( ) ln

=A e₀ rt

(b) Part (a) shows that as the number of compoundings per year increases toward infinity, the limit of interest compoundedk times per year is interest compounded continuously.

58. (a) Forx f x g x

≠ ′

′ = =

0 1

1 1

, ( )

( ) .

lim ( ) ( )

x

f x g x

→

′

′ =

0 1

lim ( ) ( )

x

f x g x

→0 = =

2

1 2

(b) This does not contradict L’Hôpital’s Rule since lim ( )

x→0f x =2 and limx→0g(x)= 1.

59. (a) A t( )=

∫

te dx−x = −_⎣⎡ e−x⎤_{⎦ =}t e−t+

0 0 1

lim ( ) lim( ) lim

t→∞A t t→∞ e t et

−

→∞

= − + = ⎛− +

⎝⎜ ⎞ ⎠ 1

1 1 1_{⎟⎟ =}1

(b) V t( )=π

∫

t(e−x)2dx 0

= −

∫

π te 2xdx

0 = ⎡−

⎣⎢ ⎤ ⎦⎥

−

π 1

2

2 0 e x

t

= ⎛− +

⎝⎜ − ⎞⎠⎟

π 1

2 1 2

2 e t

=π

(

− − +

)

2 1

2 e t

lim ( ) ( ) lim

( ) (

t t

t

V t A t

e

→∞ →∞

−

= − +

− + =

π π

2 1

1 2 1

2 ₎₎

1 = π2

(c) lim ( ) ( ) lim

( )

t t

t

V t A t

e

→ →

−

+ = +

− +

0 0

2

2 1

1

π

=

→ −

−

+

lim

( )

t

e

e 0

2

2 2

π

= =

π π

(11)

60. (a) _x _ƒ₍_x₎

0.1 0.04542

0.01 0.00495

0.00 0.00050

0.0001 0.00005 The limit appears to be 0. (b) lim sin

x

x x

→01 2+ = = 0

1 0

L’Hôpital’s Rule is not applied here because the limit is not of the form0

0or

∞

∞, since the denominator has

limit 1.

61. (a) f x( )=exln 1+1/( x)

1+ >1 0

x whenx<−1orx>0

Domain: (−∞ − ∪, 1) ( ,0 ∞) (b) The form is0−1, so lim ( )

x

f x

→−₁− = ∞

(c) lim

x

→−∞ ln 1

1 1

1

+ ⎛ ⎝⎜

⎞ ⎠⎟=

+ ⎛ ⎝⎜ ⎞⎠⎟

→−∞

x

lim

= − ⎛

⎝⎜ ⎞⎠⎟⎛⎝⎜ + ⎞⎠⎟ −

→−∞

−

lim

x

x x

x

1 1 1 1

2

1

2

=

+ =

→−∞ lim

x

1

1 1

1

lim ( ) lim ( / )

x x

f x e e

→−∞ = →−∞ =

ln 1+1

62.False. Needg′ (a) ≠ 0. Considerf x( ) sin= 2xandg(x)=x2

witha= 0. Here lim ( ) lim ( ) .

x→0f x′ =x→0g x′ =0

63.False. The limit is 1. 64.C. lim

tan sec

x

x x

→0 = 2 = =

1 1

65.D. lim lim lim /

x x x

x

x x

→ → →

−

− = = =

1 2

3 1

3 2

1 1

1

2 2 1 22

66.B. limlog

log lim

ln ln

x x

x

→∞ = →∞ =

2 3

1 2 1

3 3 2

67.E. lim lim ln

/

x

x x

x

→∞ + →∞

⎛

⎝⎜ ⎞⎠⎟ = +

⎛ ⎝⎜ ⎞⎠⎟ ⎛

⎝

1 1 1

1

1 3

3 ⎜⎜ ⎜

⎞

⎠ ⎟

⎟= lim→∞−− +

( )

/

x

x x x

1 1 1 3 2

=

+ = =

→∞ →∞

lim

( ) lim

x x

x x x

3 1

3

1 3

2

lim ln ( )

x f x

e e

→∞ =

3

68. Possible answers:

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

( ) lim

( )

lim

x x x

f x g x

x x

→ = → →

−

− =

3 3 3

7 3

3 7

1 7

(b) f x( )= −(x 3) ; ( )2 g x = −x 3 lim ( )

( ) lim

( )

lim ( )

x x x

f x g x

x x

x

→ = → →

− −

−

3 3

2 3

3 3

2 2 3

1

1 =0

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

( ) lim( ) lim (

x x x

f x g x

x

x x

→ = → →

−

− = −

3 3 3 3

3 3

1 3 3))2= ∞ 69. Answers may vary.

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

lim ( )

( ) lim lim

x x x

f x g x

x x

→∞ = →∞ + = →∞ =

3 1 3

1 3

(b) f x( )= +x 1; ( )g x =x2

lim ( )

( ) lim lim

x x x

f x g x

x

x x

→∞ = →∞ + = →∞ =

1 1

2 0

2

(c) f x( )=x2; ( )g x = +x 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−cosx6 is 0 for even moderately small values of x. For example,

atx=0 1. , cosx6≈0 9999999999995. (13 places), so on a 10-place calculator,cosx6=1and1−cosx6=0. (b) Same reason as in part (a) applies.

(c) lim cos lim sin

x x

x

→ →

− ₌

0 6

12 ₀

5 6

11

1 6

12

=

→ limsin

x

x x 0

6 6

2

=

→

lim cos

x

x x

x 0

5 6

5

6 12

= =

→ limcos

x

x 0

6

(12)

70. Continued

(d) The graph and/or table on a grapher show the value of the function to be 0 for x-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 =3x2,g x′( )=2x−1

f f g g

c c

c

( )1 ( 1) 2, ( )1 ( 1) 2 3

2 1

2 2

3 2

2

− − = − − = −

− =− = − cc c c

c c

+ + − =

− + =

1

3 2 1 0

3 1 1 0

2

( )( )

c=1 c= −

3or 1

The value of c that satisfies the property isc=1 3. (b) f′( )x = −sin ,x g x′( )=cosx

f π f g π g

2 0 1 2 0 1

⎛

⎝⎜ ⎞⎠⎟− ( )= − , ⎛⎝⎜ ⎞⎠⎟− ( )= −sin _{= −}

cos

c c

1 1 tanc=1

c=tan−11= 4

π _on ₀

2 ,π

⎛ ⎝⎜ ⎞⎠⎟

72. (a) lnf x( )g x( )=g x( )lnf x( )

lim ( ( ) ( )) lim ( ) lim ( )

x→c g x lnf x =

(

x→cg x

)

((

x→clnf x

)

= ∞ − ∞ = − ∞( ) lim ( ) ( ) lim ( ) ( )

x c g x

x c f x g 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( ) ( / )

/

x

x x

x

→

+ − −

0 4 3

2

1 4 3 1

= + −

→

lim / ( ) ( / )

/

x

x x 0

1 3

4 3 1 4 3

2

= +

→

−

lim / ( )

/

x

x 0

2 3

4 9 1

2

=2

9

2.D. lim ( )

x

→0+ 2

3

=

→+ lim ln

/

x

x x

0 1 2

= =

→+

lim /

/

x

x x

0 2

1

1 2 0

lim ln ( )

x

f x

e e

→0 = =

0

3 3 3

3. B. lim sin

x x

t dt

x

→

∫

− 2

2

2 ₄

= −

−

→

limcos cos

x

x x

2 2

2 4

= =

→

limsin sin

x

x x 2 2

2 4

4. (a) 1 2 4

1 2

1 3

/ /

− ⎛ ⎝⎜

⎞ ⎠⎟ = − −

− =

4 1 2/ 8 (b) −1

2

(c) a_n

n

n n

= ⎛− ⎝⎜

⎞

⎠⎟ = − − −

8 1

2 1 2

1 4

( ) ( )

(d) a_n= −⎛ a_n ⎝⎜ ⎞⎠⎟ −

1

2 1

Section 8.3

Relative Rates of Growth

(pp. 453–458)

Exploration 1 Comparing Rates of Growth asx→→ ∞

1. lim lim(ln )( ) lim(ln )

x x

x

x x

x

a x

a a x

a a

→∞ 2= →∞ = →∞

2

2 22 = ∞, soa

x_grows

faster thanx2as x→ ∞.

2. lim lim .

x x x _x

x

→∞ = →∞ = ∞ 3

2 1 5

3. lim lim

x x x _x

x

a b

→∞ = →∞

⎛

⎝⎜ ⎞⎠⎟ = ∞because a b > 1.

Quick Review 8.3

1. limln lim

x x x

x x

x

e e

→∞ = →∞ =

1

0

2. lim lim lim lim

x x

e x

e

(13)

3. lim

x x

x e

→−∞ = ∞

2 2

4. lim lim lim

x x x x x x

x e

x

e e

→∞ = →∞ = →∞ =

2

2 2 2

2 2

2

4 0

5. −3x4

6. 2 2

3 2 x x = x

7. lim ( ) ( ) lim

ln lim

x x x

f x g x

x x x

x

→∞ = →∞ →∞

+ ₌ 1+1₌

1 1

8. lim ( )

( ) lim lim

x x x

f x g x

x x

→∞ = →∞ →∞

+ ₌ ₊ ₌

4 5

2 1

5

4 1

2

9. (a) f x e x e

x e

x

x x

( )= + = +

2 2

1

′ = − = −

f x xe x e e

x x e

x x

( ) 2 2

2 2

2

0

2 x x

ex

− ₌

x(2− =x) 0

x x

f x x x

= =

′ < < >

0 2

0 0 2

or

for or

( )

The graph decreases, increases, and then decreases.

f f

e

( )0 =1; ( )2 = +1 4₂≈1 541.

f has a local maximum at ≈ (2, 1.541) and has a local minimum at (0, 1).

(b)f is increasing on [0, 2]

(c)f is decreasing on (− ∞, 0] and [2, ∞). 10. f x x x

x

x x x

( )= +sin = +1 sin , ≠0

Observe that sinx .

x <1since sinx< xforx≠0

lim ( ) limsin

x f x x

x x

→0 = +1 →0 = + =1 1 2

Thus the values of f get close to 2 as x gets close to 0, so f

doesn’t have an absolute maximum value. f is not defined at 0.

1. lim lim lim li

x

x x

e x x

e x

e

→∞ 3_{− +}₃ ₁= →∞₃ 2₋₃= →∞₆ =_x mm

x

e

→∞6 = ∞

2. lim lim !

x x

e x

e

→∞ 20 = →∞₂₀ = ∞

3. lim lim

sin , cos ,

cos cos

x x

x _x

x x

e e

e

x e x

→∞ = →∞₋ − ≤1 ≤1 llim

sin

x

x y

e x e

→∞₋ cos = ∞

4. lim

( / ) !( / )

x x

e e

x

→∞ 5 2 = 5 2 = ∞

5. lim ln

ln lim

/

/ lim

x x x

x

x x x

→∞ − = →∞ − = →∞ ₋ = 1

1

1 0

2

6. lim ln lim /

/ ( ) / ( ) /

x x

x

x x x

→∞ = →∞ − = − =

1 1 2

1 2 1 2 00

7. lim ln lim /

/ ( ) / lim / (

x x x

x x

→∞3 = →∞ −2 3= →∞ 1

1 3

1

1 3 xx)−2 3/ x=0

8. lim ln lim / lim

x x x

x x

x

x x

→∞ 3 = →∞ 2 = →∞ 3=

1 3

1

3 0

9. lim lim lim

x x x

x x

x

x x

→∞ →∞ →∞

+ ₌ _{+ =} ₌

2 2

4 2 4

2

2 1

10. lim lim lim

x x x

x x

x

x x

x

x x

→∞ →∞ →∞

+

= + = =

4 2

4 4

2 2

5 5 12

12 11

11. lim( ) lim lim

/

x x x

x

→∞ →∞ →∞

+ ₌ + ₌

6 2 1 3 2

6 2

6

120 33

3

120x =1

12. lim sin lim cos , cos , l

x x

x

x x

→∞ →∞

+ ₌ + _{− ≤} _≤

2 2

2

2 1 1 iimx

x x

→∞ = 2

2 1

13. lim log ln

ln

/ ln

x

x x

→∞ = =

1 1

2 10

1

1 2 10

14. lim

x x

x

e

e e

→∞ +

= 1

15.First observe that 1+x4grows at the same rate as x2.

lim lim lim

x x x

x x

x

x x

→∞ →∞ →∞

+

= = + = + =

1 1 1

1 1

4 2

4

4 4

Next comparex2withex.

lim lim lim

x x x x x x

x e

x

e e

→∞ = →∞ = →∞ =

2 ₂ ₂

0

x2 ex x x4

1 grows slower than as → ∞,so + grows slower thanexasx→ ∞.

16. lim lim .

x x x _x

x

e e e

→∞ = →∞

⎛

⎝⎜ ⎞⎠⎟ = ∞ >

4 4 4

1 since

(14)

17. lim ln lim

ln

x x x x

x x x e

x

x x

e

→∞ − = →∞

⎛ ⎝⎜

⎞ ⎠⎟+ −

1

=

→∞ limln

x x

x e

= =

→∞ lim /

x x

x e

1 0

xlnx− x grows slower thanexasx→ ∞.

18. lim lim

x x

x _x

xe

e x

→∞ = →∞ = ∞

xexgrows faster thanexasx→ ∞.

19. lim

x x

x e

→∞ =

1000

0 Repeated application of L'Hôpitaal's

Rule gets

⎛ ⎝⎜

= ⎞ ⎠⎟

→∞

lim ! .

x _ex

1000 0

x1000grows slower thanexasx→ ∞.

20. lim( ) / lim

x

x x

x _x x

e e

→∞ −

→∞

+ ₌ ⎛ ₊

⎝⎜ ⎞⎠⎟=

2 1

2 1 2

1 2

2

ex+e−x

2 grows at the same rate ase x

x

as → ∞.

21. lim lim

x x

x

x x x

→∞ + = →∞ +

⎛ ⎝⎜

⎞ ⎠⎟= ∞ 3

2 2

3 3

x3+ 3 grows faster than x2asx→ ∞.

22. lim lim

x x

x

x x x

→∞ + = →∞ +

⎛

⎝⎜ ⎞⎠⎟=

15 3 15 3

0

2 2

15x+ 3 grows slower thanx2asx→ ∞.

23. lim ln lim / lim

x x x

x x

x

x x

→∞ 2 = →∞ = →∞ 2=

1 2

1

2 0

lnx grows slower thanx2asx→ ∞.

24. lim lim(ln ) lim(ln )

x x

x

x x

x

x x

→∞ = →∞ = →∞ =

2 2 2

2

2 2 2

2

2 ∞ ∞.

2xgrows faster than x2asx→ ∞.

25. limlog

ln lim

log

ln lim

(ln

x x x

x x

x

→∞ = →∞ = →∞

2 2

2

2 2 )) / (ln )

ln ln

2 2

2

x =

log₂x2grows at the same rate as ln x as x→ ∞.

26. lim /

ln lim ln

x x

x

x x x

→∞ = →∞ =

1 1

0

1

xgrows slower thanlnxasx→ ∞.

27. lim

ln lim ln

x x

e

x e x

→∞ −

→∞

= 1 =0

e−xgrows slower thanlnxasx→ ∞.

28. lim ln ln

x

x x

→∞ =

5 5

5lnx grows at the same rate as lnxasx→ ∞, 29. Compareextoxx.

lim lim

x x x _x

x

e x

→∞ = →∞

⎛ ⎝⎜

⎞ ⎠⎟ =0 exgrows slower thanxx. Compareexto(ln ) .x x

lim

(ln ) lim ln

x x

x _x

x

e x

→∞ = →∞

⎛ ⎝⎜

⎞ ⎠⎟ =0

exgrows slower than(ln ) .x x

Comparex exto ex/2.

lim _/ lim /

x x

x _x

x

e

e e

→∞ 2= →∞ = ∞

2

exgrows faster thanex/2. Compare xxto(ln ) .x x

lim

(ln ) lim ln

x x

x _x

x

x x

→∞ = →∞

⎛ ⎝⎜

⎞

⎠⎟ = ∞since lim→→∞ = →∞ = ∞

x x

x

ln lim1/ .

xxgrows faster than(ln ) .x x

Thus, in order from slowest-growing to fastest-growing, we getex/2, ex, (ln ) ,xx xx.

30.Compare 2xtox2

lim lim(ln ) lim(ln )

x x

x

x x

x

x x

→∞ = →∞ = →∞ =

2 2 2

2

2 2 2

2

2 ∞ ∞

2xgrows faster thanx2. Compare2xto (ln 2)x.

lim

(ln ) lim ln

x x

x _x

x

→∞ = →∞

⎛ ⎝⎜

⎞ ⎠⎟ = ∞

2 2

2

2 since

2 ln 22>1. 2xgrows faster than(ln )2x

Compare 2xtoex.

lim lim .

x x x _x

x

e e e

→∞ = →∞

⎛

⎝⎜ ⎞⎠⎟ = <

2 2

0since2 1