solucionario calculo de una variable - 9° ediciÓn - ron larson

LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS GRATIS EN

DESCARGA DIRECTA

SIGUENOS EN:

VISITANOS PARA DESCARGARLOS GRATIS.

C H A P T E R 1Limits and Their Properties

Section 1.1 A Preview of Calculus . . . . . . . . . . . . . . . . . . . 305

Section 1.2 Finding Limits Graphically and Numerically . . . . . . . 305

Section 1.3 Evaluating Limits Analytically . . . . . . . . . . . . . . . 309

Section 1.4 Continuity and One-Sided Limits . . . . . . . . . . . . . 315

Section 1.5 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . 320

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327

305


Section 1.1 A Preview of CalculusSolutions to Even-Numbered Exercises

2. Calculus: velocity is not constantDistance 20 ftsec15 seconds 300 feet

4. Precalculus: rate of change slope 0.08

6. Precalculus: 2

Area 22 8. Precalculus: Volume 326 54

10. (a)

(b) You could improve the approximation by using more rectangles.

Area 125 5

1.5 52

52.5

53

53.5

54

54.5 9.145

Area 5 52 53

54 10.417

Section 1.2 Finding Limits Graphically and Numerically

2.

Actual limit is 14 .limx2

x 2x2 4 0.25

x 1.9 1.99 1.999 2.001 2.01 2.1

0.2564 0.2506 0.2501 0.2499 0.2494 0.2439f x

4.

14 .Actual limit islim

x3

1 x 2x 3 0.25

x 3.1 3.01 3.001 2.999 2.99 2.9

0.25160.25020.25000.25000.24980.2485f x

6.


xx 1 45x 4 0.04

x 3.9 3.99 3.999 4.001 4.01 4.1

0.0408 0.0401 0.0400 0.0400 0.0399 0.0392f x

8.

(Actual limit is 0.) (Make sure you use radian mode.)limx0

cos x 1x

0.0000

x 0.1 0.01 0.001 0.001 0.01 0.1

0.0500 0.0050 0.0005 0.0005 0.0050 0.0500f x

10. limx1

x2 2 3 12. limx1

f x limx1

x2 2 3

14. does not exist since the

function increases and decreaseswithout bound as x approaches 3.

limx3

1x 3 16. limx0 sec x 1 18. limx1 sinx 0

20.

(a)

(b)

(c)

does not exist. The values of C jump from to at t 3.0.710.59limt3.5

Ct

limt3.5

Ct 0.71

00

5

1

Ct 0.35 0.12t 1

t 3 4

0.710.710.710.710.710.710.59Ct

3.73.63.53.43.3

t 3 3 4

0.710.710.710.590.590.590.47Ct

3.53.12.92.5

22. You need to find such that impliesThat is,

So take

Then implies

Using the first series of equivalent inequalities, you obtain

f x 3 x2 4 < 0.2.

3.8 2 < x 2 < 4.2 2.

4.2 2 < x 2 < 4.2 20 < x 2 <

4.2 2 0.0494.

0.2 0:limx4

x 4 2 36.

Given

Hence, let

Hence for you have

f x L <

x 3 0 <

x 3 < 0 < x 3 < , .

x 3 <

x 3 0 <

> 0:

limx3

x 3 0

Section 1.2 Finding Limits Graphically and Numerically 307

38.

Given

If we assume then

Hence for you have

f x L < x2 3x 0 <

xx 3 < x 3 < 14 0:

limx3

x2 3x 0 40.

The domain is all The graphing utility does notshow the hole at 3, 12.

x 1, 3.

limx3

f x 123 5

4

4

f x x 3x2 4x 3

42.

The domain is all The graphing utility does notshow the hole at 3, 16.

x 3.

limx3

f x 169 3

3

3f x x 3x2 9 44. (a) No. The fact that has no bearing on the exis-tence of the limit of as x approaches 2.

(b) No. The fact that has no bearing on thevalue of f at 2.

limx2

f x 4f x

f 2 4

46. Let be the atmospheric pressure in a plane ataltitude x (in feet).

limx0

px 14.7 lbin2

px 48.

Using the zoom and trace feature, That is, for

x2 4x 2 4 < 0.001.0 < x 2 < 0.001, 0.001.

1.998 2.0020

(1.999, 0.001)(2.001, 0.001)

0.002

50. True 52. False; let

and f 4 10 0limx4

f x limx4

x2 4x 0

f x x2 4x,10,x 4x 4

.

54. limx4

x2 x 12x 4 7

n

1 4.1 7.1

2 4.01 7.01

3 4.001 7.001

4 4.0001 7.0001

f 4 0.1n4 0.1n n1 3.9 6.9

2 3.99 6.99

3 3.999 6.999

4 3.9999 6.9999

f 4 0.1n4 0.1n

308 Chapter 1 Limits and Their Properties

Section 1.3 Evaluating Limits Analytically

56. Let be given. Take

If then

which shows that limxc

mx b mc b.

mx b mc b < mx mc < mx c <

0 < x c < m,

m. > 0f x mx b, m 0. 58. Let There exists such that implies

That is,

Hence for x in the interval

gx > 12L > 0.

c , c , x c,

12L 0.

2. (a)(b)

gx 12x 3x 9

limx0

gx 4

limx4

gx 2.4

0 10

5

10 4. (a)(b)

f t tt 4

limt1

f t 5limt4

f t 0

5 10

10

10

6. limx2

x3 23 8 8. limx3

3x 2 33 2 7

10. limx1

x2 1 12 1 0 12. limx1

3x3 2x2 4 313 212 4 5

18. limx3

x 1x 4

3 13 4 2 20. limx4

3x 4 34 4 2

22. limx0

2x 13 20 13 1 24. (a)(b)(c) lim

x3 g f x g4 16

limx4

gx 42 16

limx3

f x 3 7 4

14. limx3

2x 2

23 2 2 16. limx3

2x 3x 5

23 33 5

38

26. (a)(b)(c) lim

x4 g f x g21 3

limx21

gx 321 6 3

limx4

f x 242 34 1 21 28. limx

tan x tan 0

30. limx1

sin x2 sin

2 1 32. limx cos 3x cos 3 1

34. limx53

cos x cos 53

12 36. limx7 sec

x

6 sec 76

233

Section 1.3 Evaluating Limits Analytically 309

38. (a)

(b)

(c)

(d) limxc

f xgx

limxc

f xlimxc

gx 3212 3

limxc

f xgx limxc f x limxc gx 32

12

34

limxc

f x gx limxc

f x limxc

gx 32 12 2

limxc

4f x 4 limxc

f x 432 6 40. (a)(b)

(c)

(d) limxc

f x23 limxc

f x23 2723 9

limxc

f x2 limxc

f x2 272 729

limxc

f x18

limxc

f xlimxc

18 2718

32

limxc

3f x 3limxc f x 327 3

42. and agree except at

(a)

(b) limx0

hx limx0

f x 3

limx2

hx limx2

f x 5

x 0.hx x2 3x

xf x x 3 44. and agree except at

(a) does not exist.

(b) limx0

f x 1

limx1

f x

x 0.f x xx2 x

gx 1x 1


4

8

8

4

limx1

f x limx1

gx 5

x 1.

gx 2x 3f x 2x2 x 3x 1 48. and agree except at

4 4

1

7

limx1

f x limx1

gx 3

x 1.

gx x2 x 1f x x3 1

x 1

50.

limx2

1x 2

14

limx2

2 xx2 4 limx2

x 2x 2x 2 52.

limx4

x 1x 2

36

12

limx4

x2 5x 4x2 2x 8 limx4

x 4x 1x 4x 2

54.

limx0

2 x 22 x 2x limx0

12 x 2

1

22

24

limx0

2 x 2x

limx0

2 x 2x

2 x 22 x 2

56. limx3

x 1 2x 3 limx3

x 1 2x 3

x 1 2x 1 2

limx3

x 3x 3x 1 2 limx3

1x 1 2

14

58. limx0

1x 4

14

x lim

x0

4 x 44x 4

x lim

x0

14x 4

116

60. lim

x0

x x2 x2

x

lim

x0

x2 2xx x2 x2

x lim

x0

x2x x

x lim

x02x x 2x


62.

lim

x0

x3x2 3xx x2

x

lim

x0

3x2 3xx x2 3x2

lim

x0

x x3 x3

x

lim

x0

x3 3x2x 3xx2 x3 x3

x

64.

Analytically,

. limx16

1x 4

18

limx16

4 xx 16 limx16

4 xx 4x 4

f x 4 xx 16

x 15.9 15.99 15.999 16 16.001 16.01 16.1

.1252 .125 .125 ? .125 .125 .1248f xIt appears that the limit is 0.125.

0 20

1

1

66.

Analytically,

.

(Hint: Use long division to factor )x5 32.

limx2

x4 2x3 4x2 8x 16 80

limx2

x5 32x 2 limx2

x 2x4 2x3 4x2 8x 16x 2

4 3

25

100limx2

x5 32x 2 80

x 1.9 1.99 1.999 1.9999 2.0 2.0001 2.001 2.01 2.1

72.39 79.20 79.92 79.99 ? 80.01 80.08 80.80 88.41f x

68. 30 0limx0

31 cos xx

limx0

31 cos xx 70. lim0 cos tan

lim

0

sin

1

72.

10 0

limx0

tan2 xx

limx0

sin2 xx cos2 x

limx0

sin xx sin x

cos2 x 74. lim sec 1

76.

2

limx4

sec x

limx4

1cos x

limx4

sin x cos xcos xsin x cos x

limx4

1 tan xsin x cos x limx4

cos x sin xsin x cos x cos2 x

78. 21131 23limx0

sin 2xsin 3x limx0 2

sin 2x2x

13

3xsin 3x


80.

Analytically, .limh0

1 cos 2h 1 cos0 1 1 2

5 5

4

4f h 1 cos 2h

h 0.1 0.01 0.001 0 0.001 0.01 0.1

1.98 1.9998 2 ? 2 1.9998 1.98f h

82.

Analytically, .limx0

sin x3x

limx0

3x2sin xx 01 0

3 3

2

2f x sin x3x

x 0.1 0.01 0.001 0 0.001 0.01 0.1

0.215 0.0464 0.01 ? 0.01 0.0464 0.215f x

The limit appear to equal 0.


84.

limh0

1x h x

1

2x lim

h0

x h xhx h x

limh0

x h xh

x h xx h x

limh0

f x h f xh limh0

x h xh

86.

limh0

h2x h 4h limh0 2x h 4 2x 4

limh0

f x h f xh limh0

x h2 4x h x2 4xh limh0

x2 2xh h2 4x 4h x2 4xh

88.

Therefore, limxa

f x b. b lim

xa f x b

limxa

b x a limxa

f x limxa

b x a 90.

limx0

x sin x 0

2

2 2

6

f x x sin x

92.

limx0

x cos x 0

6

2 2

6

f x x cos x 94.

limx0

x cos 1x 0

0.5 0.5

0.5

0.5

hx x cos 1x


96. and agree at all points

except x 1.

gx x 1f x x2 1

x 1 98. If a function f is squeezed between two functions h and g,and h and g have the same limit L as

then exists and equals L.limxc

f xx c,hx f x gx,

100.

3 3

2

g

hf

2

hx sin2 x

xgx sin2 x,f x x,

When you are close to 0 the magnitude of g is smallerthan the magnitude of f and the magnitude of g isapproaching zero faster than the magnitude of f.Thus, when x is close to 0g f 0

102. when seconds

limt5102

16t 5102 8010 ftsec 253 ftsec

limt5102

16t2 1252

510

2 t

limt5102

16t 5102 t 510

2

t 5102

limt5102

s5102 st510

2 t

limt5102

0 16t2 1000510

2 t

t 100016 5102st 16t2 1000 0

104. when seconds.

The velocity at time is

Hence, if the velocity is 9.8150049 54.2 msec.a 150049,

limta

4.9a t 2a4.9 9.8a msec.

limta

sa sta t

limta

4.9a2 150 4.9t2 150a t

limta

4.9a ta ta t

t a

t 1504.9 150049 5.534.9t2 150 0

106. Suppose, on the contrary, that exists. Then, since exists, so would which is acontradiction. Hence, does not exist.lim

xc gx

limxc

f x gx,limxc

f xlimxc

gx

108. Given n is a positive integer, then

cclimxc

xx n3 . . . c n.

c limxc

xx n2 climxc

xlimxc

x n2

limxc

x n limxc

xx n1 lim xc

xlimxc

x n1f x x n,

110. Given

For every there exists such that whenever

Now for < Therefore, limxc

f x 0..x c f x 0 < f x 0 f x 0 < x c < . f x 0 < > 0 > 0,

limxc

f x 0:


112. (a) If then

Therefore,

(b) Given For every there exists such that whenever Since for then lim

xc f x L.x c < ,< f x L f x L

0 < x c < . f x L < > 0 > 0,limxc

f x L:limxc

f x 0. 0 lim

xc f x 0

limxc

f x limxc

f x limxc

f x f x f x f x

limxc

f x 0.limxc

f x 0,

114. True. limx0

x3 03 0 116. False. Let

Then but f 1 1.limx1

f x 1

f x x 3 x 1x 1 , c 1

118. False. Let and Then for all But lim

x0 f x lim

x0 gx 0.x 0.

f x < gxgx x2.f x 12 x2 120.

10 0

limx0

sin xx limx0

sin x1 cos x

limx0

sin xx

sin x

1 cos x

limx0

1 cos2 xx1 cos x limx 0

sin2 xx1 cos x

limx0

1 cos xx

limx0

1 cos xx

1 cos x1 cos x

122.

(a) The domain of f is all (b)

The domain is not obvious. The hole at is notapparent.

x 0

2

23

23

2

2 n.x 0,

f x sec x 1x2

(c)

(d)

Hence,

1112 12.

limx0

sec x 1x2

limx0

1cos2 x

sin2 xx2

1sec x 1

tan2 x

x2sec x 1 1

cos2 xsin2 x

x2

1

sec x 1

sec x 1x2

sec x 1

x2

sec x 1sec x 1

sec2 x 1x2sec x 1

limx0

f x 12

124. The calculator was set in degree mode, instead of radian mode.


Section 1.4 Continuity and One-Sided Limits

2. (a)

(b)

(c)

The function is continuous atx 2.

limx2

f x 2

limx2

f x 2

limx2

f x 2 4. (a)

(b)

(c)

The function is NOT continuous atx 2.

limx2

f x 2

limx2

f x 2

limx2

f x 2 6. (a)

(b)

(c) does not exist.


limx1

f x

limx1

f x 2

limx1

f x 0

8. limx2

2 xx2 4 limx2

1x 2

14

10.

limx4

1x 2

14

limx4

x 4x 4x 2

limx4

x 2x 4 limx4

x 2x 4

x 2x 2

12. limx2

x 2x 2 limx2

x 2x 2 1

14.

2x 0 1 2x 1

limx0

2x x 1

limx0

2xx x2 xx

limx0

x x2 x x x2 xx

limx0

x2 2xx x2 x x x2 xx

16.

limx2

f x 2

limx2

f x limx2

x2 4x 6 2

limx2

f x limx2

x2 4x 2 2 18. limx1

f x limx1

1 x 0

20. does not exist since

and do not exist.limx2

sec xlimx2

sec x

limx2

sec x 22. limx2

2x x 22 2 2

24. limx1

1 x2 1 1 2 26.has a discontinuity at since is not defined.f 1x 1

f x x2 1

x 1

28. has discontinuity at since f 1 2 limx1

f x 1.x 1f x x,2,2x 1,x < 1x 1x > 1

30. is continuous on 3, 3.f t 3 9 t2 32. is not defined. g is continuous on 1, 2.g2

Section 1.4 Continuity and One-Sided Limits 315

34. is continuous for all real x.f x 1x2 1 36. is continuous for all real x.f x cos

x

2

38. has nonremovable discontinuities at and since and do not exist.limx1

f xlimx1

f xx 1x 1f x xx2 1

40. has a nonremovable discontinuity at since does not exist, and has a removable discontinuity

at since

.limx3

f x limx3

1x 3

16

x 3

limx3

f xx 3f x x 3x2 9

42.

has a nonremovable discontinuity at sincedoes not exist, and has a removable discontinu-

ity at since

limx1

f x limx1

1x 2

13.

x 1lim

x2 f x

x 2

f x x 1x 2x 1 44.has a nonremovable discontinuity at since does not exist.

limx3

f xx 3f x x 3

x 3

46.

has a possible discontinuity at .

1.

2.

3.

f is continuous at therefore, f is continuous for all real x.x 1,

f 1 limx1

f x

limx1

f x 1limx1 f x limx1 2x 3 1lim

x1f x lim

x1x2 1

f 1 12 1x 1

f x 2x 3,x2, x < 1x 1

48. has a possible discontinuity at

1.

2. does not exist.

Therefore, f has a nonremovable discontinuity at x 2.

limx2

f xlimx2

f x limx2

2x 4lim

x2f x lim

x2x2 4x 1 3

f 2 22 4

x 2.f x 2x,x2 4x 1, x 2x > 2

50. has possible discontinuities at

1.

2.

3.

f is continuous at and therefore, is continuous for all real x.fx 5,x 1

f 5 limx5

f xf 1 limx1

f x

limx5

f x 2limx1

f x 2

f 5 csc 56 2f 1 csc

6 2

x 5.x 1, csc x6 ,

2,1 x 5x < 1 or x > 5

f x csc x6

,

2,x 3 2x 3 > 2


52. has nonremovable discontinuities at each

k is an integer.2k 1,

f x tan x254. has nonremovable discontinuities at each

integer k.f x 3 x

56.

f is not continuous at x 4

limx0

fx 08 8

10

20limx0

f x 0 58.

Let a 4.

limx0

gx limx0

a 2x a

limx0

g(x limx0

4 sin xx

4

60.

Find a such that 2a 8 a 4.

limxa

x a 2a

limxa

gx limxa

x2 a2

x a

62.

Nonremovable discontinuity at Continuous for all .Because is not defined for it is better to say that is discontinuous from the right at x 1.f gx < 1,f g

x > 1x 1.

f gx 1x 1

64.

Continuous for all real x

f gx sin x2 66.

Nonremovable discontinuity at and

3 4

2

2

x 2.x 1

hx 1x 1x 2

68.

Therefore, and f is continuous on the entire real line. ( was the only possible discontinuity.)x 0limx0

f x 0 f 0lim

x0 f x lim

x0 5x 0

limx0

f x limx0

cos x 1x

0

f 0 50 0 7

3

2

3

f x cos x 1

x,

5x,

x < 0

x 0

70.

Continuous on 3,

f x xx 3 72.

Continuous on 0,

f x x 1x


74.

The graph appears to be continuous on the intervalSince is not defined, we know that f has

a discontinuity at This discontinuity is removableso it does not show up on the graph.

x 2.f 24, 4.

4 40

14

f x x3 8

x 2 76. is continuous on

and

By the Intermediate Value Theorem, for at leastone value of c between 0 and 1.

f x 0f 1 2f 0 2

0, 1.f x x3 3x 2

78. is continuous on

and

By the Intermediate Value Theorem, for at leastone value of c between 1 and 3.

f 1 0

f 3 43 tan 38 > 0.f 1 4 tan

8 < 0

1, 3.f x 4x

tan x

8 80.

is continuous on

and

By the Intermediate Value Theorem, for at leastone value of c between 0 and 1. Using a graphing utility,we find that x 0.5961.

f x 0f 1 2f 0 2

0, 1.f xf x x3 3x 2

82.

h is continuous on

and

By the Intermediate Value Theorem, for at leastone value between 0 and 1. Using a graphing utility, wefind that 0.4503.

h 0

h1 2.67 < 0.h0 1 > 0

0, 1.

h 1 3 tan 84.

f is continuous on and

The Intermediate Value Theorem applies.

( is not in the interval.)Thus, f 2 0.

x 4c 2

x 2 or x 4

x 2x 4 0

x2 6x 8 0

1 < 0 < 8

f 3 1f 0 80, 3.

f x x2 6x 8

86.

f is continuous on The nonremovable discontinuity,lies outside the interval.

and

356 < 6 <

203

f 4 203f 52

356

x 1,52 , 4.

f x x2 x

x 1 The Intermediate Value Theorem applies.


x 2c 3

x 2 or x 3

x 2x 3 0

x2 5x 6 0

x2 x 6x 6

x2 x

x 1 6


88. A discontinuity at is removable if you can define(or redefine) the function at in such a way that thenew function is continuous at Answers will vary.

(a)

(b) f x sinx 2x 2

f x x 2x 2

x c.x c

x c

(c)

1 2 313 21

2

3

1

3

2

y

x

f x 1,0,1,0,

if x 2if 2 < x < 2if x 2if x < 2

90. If f and g are continuous for all real x, then so is (Theorem 1.11, part 2). However, might not be continuous if For example, let and Then and are continuous for all real x, but is not continuous at x 1.fggfgx x2 1.f x x

gx 0.fgf g

92.

Nonremovable discontinuity at each integer greater than 2.

C 1.04,1.04 0.36t 1,1.04 0.36t 2,0 < t 2t > 2, t is not an integert > 2, t is an integer

You can also write C as

.

t1 2 3 4

1

2

3

4

C

C 1.04,1.04 0.362 t, 0 < t 2t > 2

94. Let be the position function for the run up to the campsite. ( corresponds to 8:00 A.M., (distanceto campsite)). Let be the position function for the run back down the mountain: Let When (8:00 A.M.), .When (8:10 A.M.), .Since and then there must be a value t in the interval such that If then

which gives us Therefore, at some time t, where 0 t 10, the position functions for therun up and the run down are equal.

st rt. 0,st rtf t 0,f t 0.0, 10f 10 > 0,f 0 < 0

f 10 s10 r10 > 0t 10f 0 s0 r0 0 k < 0t 0

f t st rt.r10 0.r0 k,rts20 kt 0s0 0st

96. Suppose there exists in such that > 0 and there exists in such that Then by the IntermediateValue Theorem, must equal zero for some value of x in (or ). Thus, f would have a zero in ,which is a contradiction. Therefore, > 0 for all x in or for all x in .a, bf x < 0a, bf x

a, bx2, x1 if x2 < x1x1, x2f xf x2 < 0.a, bx2f x1a, bx1

98. If then and . Hence, f iscontinuous at

If then for x rational, whereasfor x irrational. Hence, f is not

continuous for all x 0.limt x

f t limt x

kt kx 0limtx

f t 0x 0,x 0.

limx0

f x 0f 0 0x 0, 100. True1. is defined.

2. exists.

3.

All of the conditions for continuity are met.

f c limxc

f x

limxc

f x Lf c L


Section 1.5 Infinite Limits

102. False; a rational function can be written as where P and Q are polynomials of degree m and n,respectively. It can have, at most, n discontinuities.

PxQx 104. (a)

(b) There appears to be a limiting speed and a possiblecause is air resistance.

t

20

30

10

40

50

60

105 15 20 25 30

S

106. Let y be a real number. If then If then let such that (this is possiblesince the tangent function increases without bound on ). By the Intermediate Value Theorem, iscontinuous on and , which implies that there exists x between 0 and such that The argumentis similar if y < 0.

tan x y.x00 < y < M0, x0f x tan x0, 2

M tan x0 > y0 < x0 < 2y > 0,x 0.y 0,

108. 1. is defined.

2. exists.

3.

Therefore, f is continuous at x c.limxc

f x f c.Let x c x. As x c, x0

limxc

f x limx0

f c x f cf c

110. Define Since and are continuous on so is f.and

By the Intermediate Value Theorem, there exists c in such that

f c f2c f1c 0 f1c f2cf c 0.a, b

f b f2b f1b < 0.f a f2a f1a > 0a, b,f2f1f x f2x f1x.

2.

limx2

1x 2

limx2

1x 2 4.

limx2

sec x

4

limx2

sec x

4

6.

limx3

f x

limx3

f x

f x xx2 9

x

499.9 49.92 4.915 0.9091500.150.085.0821.077f x2.52.92.992.9993.0013.013.13.5


8.

limx3

f x

limx3

f x

f x sec x6

x

191.0 19.11 3.86419101910191.019.113.864f x2.52.92.992.9993.0013.013.13.5

10.

Therefore, is a vertical asymptote.x 2

limx2

4x 23

limx2

4x 23

12.

Therefore, is a vertical asymptote.


limx1

2 xx21 x

limx1

2 xx21 x

x 0

limx0

2 xx21 x

limx0

2 xx21 x

14. No vertical asymptote since the denominator is never zero. 16. and


and

Therefore, is a vertical asymptote.s 5

lims5

hs .lims5

hs

s 5

lims5

hs .lims5

hs

18. has vertical asymptotes at

n any integer.x 2n 12 ,

f x sec x 1cos x

20.

No vertical asymptotes. The graph has holes at and x 4.

x 2

x 2, 4

16 x,

gx 12x3 x2 4x

3x2 6x 24 16

xx2 2x 8x2 2x 8

22.

Vertical asymptotes at and The graph has holes at and x 2.x 3x 3.x 0

f x 4x2 x 6

xx3 2x2 9x 18 4x 3x 2xx 2x2 9

4xx 3, x 3, 2

24.

has no vertical asymptote since

limx2

hx limx2

x 2x2 1

45.

hx x2 4

x3 2x2 x 2 x 2x 2x 2x2 1 26.

Vertical asymptote at The graph has a hole att 2.

t 2.

ht tt 2t 2t 2t2 4 t

t 2t2 4, t 2

Section 1.5 Infinite Limits 321


n any integer.

There is no vertical asymptote at since

lim0

tan

1.

0

2n 1

2

2 n,

g tan

sin

cos 30.

Removable discontinuity at x 1

3 3

12

2

limx1

x2 6x 7x 1 limx1 x 7 8

32.

Removable discontinuity atx 1

3 3

2

2limx1

sinx 1x 1 1 34. limx1

2 x1 x

36. limx4

x2

x2 16 12 38. limx12

6x2 x 14x2 4x 3 limx12

3x 12x 3

58

40. limx3

x 2x2

19 42. limx0 x2

1x

44. 2cos x

limx2

46. limx0

x 2cot x

limx0

x 2tan x 0

48. and

Therefore, does not exist.limx12

x2 tan x

limx12

x2 tan x .limx12

x2 tan x 50.

8 8

4

4

limx1

f x limx1

x 1 0

f x x3 1

x2 x 1

52.

9 9

6

6

limx3

f x

f x sec x6 54. The line is a vertical asymptote if the graph of fapproaches as x approaches c.

x c

56. No. For example, has no

vertical asymptote.

f x 1x2 1 58.

(In this case we know that k > 0.)limV0

kV k

P kV


60. (a)

(b)

(c) lim2

50 sec2

r 50 sec2 3 200 ftsec

r 50 sec2 6 200

3 ftsec62.

limvc

m limvc

m0

1 v2c2

m m0

1 v2c2

64. (a)

Domain: x > 25

25xx 25 y

50x 2yx 25

50x 2xy 50y

50y 50x 2xy

50 2xyy x

50 2ddx dy

Average speed Total distanceTotal time(b)

(c)

As gets close to 25 mph, becomes larger and larger.yx

limx25

25xx 25

x 30 40 50 60

y 150 66.667 50 42.857

66. (a)

Domain:

(c)

00

1.5

100

0, 2 50 tan 50

A 12bh 12r

2 121010 tan

1210

2 (b)

(d) lim2

A

0.3 0.6 0.9 1.2 1.5

0.47 4.21 18.0 68.6 630.1f

68. False; for instance, let

The graph of f has a hole at not a verticalasymptote.

1, 2,

f x x2 1

x 1 .

70. True

72. Let and and

and but

limx0

1x2 1x4 limx0

x2 1x4 0.

limx0

1x4

, limx0

1x2

c 0.gx 1x4

,f x 1x2

74. Given then

by Theorem 1.15.

limx c

gxfx 0limx c f x , let g x 1.


Review Exercises for Chapter 1

2. Precalculus. L 9 12 3 12 8.25

4.

limx0

f x 0.2

1

0.5

1

0.5x 0.1 0.01 0.001 0.001 0.01 0.1

0.358 0.354 0.354 0.354 0.353 0.349f x

6. (a) does not exist.limx2

gxgx 3xx 2

(b) limx0

gx 0

8.

Let be given. We need

x 9 < x 3 x 3 < x 3x 3 < x 3

> 0

limx9

x 9 3. Assuming you can choose

Hence, for you have

f x L < x 3 <

x 9 < 5 < x 30 < x 9 < 5,

5.4 < x < 16,

10. Let be given. can be any positive

number. Hence, for you have

f x L < 9 9 <

0 < x 5 < , > 0lim

x5 9 9. 12. lim

y4 3y 1 34 1 9

14. limt3

t 2 9t 3 limt3 t 3 6 16.

limx0

14 x 2

14

limx0

4 x 2x

limx0

4 x 2x

4 x 24 x 2

18.

lims0

11 s 1s11 s 1 lims0

11 s11 s 1

12

lims0

11 s 1s

lims0

11 s 1s 11 s 111 s 1

20.

4

12 13

limx2

x 2x2 2x 4

limx2

x2 4x3 8 limx2

x 2x 2x 2x2 2x 4

22. limx4

4xtan x

44

1


24.

0 01 0

limx0

cos x 1x limx0 sin sin x

x

limx0

cos x 1x

limx0

cos cos x sin sin x 1x

26. limxc

f x 2gx 34 223 712

28.

(a)

Actual limit is

(c)

13

limx1

11 3x 3x2

limx1

1 xx 11 3x 3x2

limx1

1 3xx 1 limx1

1 3xx 1

1 3x 3x21 3x 3x2

13 .limx1

1 3xx 1 0.333

f x 1 3x

x 1

(b)

3

3

3

2x 1.1 1.01 1.001 1.0001

0.33330.33320.33220.3228f x

30.

When the velocity is approximately

limt6.39

4.96.39 6.39 62.6 msec.

limta

sa sta t

limta

4.9a t

t 6.39,

st 0 4.9t2 200 0 t2 40.816 t 6.39 sec

32. does not exist. The graph jumps from 2 to 3at x 4.

limx4

x 1 34. limx1

gx 1 1 2.

36. lims2

f s 2 38.

Removable discontinuity at Continuous on , 1 1,

x 1

limx1

3x 2 5 0

limx1

f x limx1

3x2 x 2x 1

f x 3x2 x 2

x 1,

0,

x 1

x 1

Review Exercises for Chapter 1 325

40.

Nonremovable discontinuity at Continuous on , 2 2,

x 2

limx2

2x 3 1

limx2

5 x 3

f x 5 x,2x 3, x 2x > 2 42.

Domain:


x 0

, 1, 0,

limx0

1 1x

f x x 1x 1 1x

44.


x 1

limx1

x 12x 1

12

f x x 12x 2 46.Nonremovable discontinuities when

Continuous on

for all integers n.

2n 14 , 2n 1

4

x 2n 1

4

f x tan 2x

48.

Find b and c so that and

Consequently we get

Solving simultaneously, b 3 and c 4.

1 b c 2 and 9 3b c 4.

limx3

x2 bx c 4.limx1

x2 bx c 2

limx3

x 1 4

limx1

x 1 2

50.

C has a nonremovable discontinuity at each integer.

00

5

30

9.80 2.50x 1

C 9.80 2.50x 1, x > 0

54.

Vertical asymptotes at and x 2x 2

hx 4x4 x2 56.

Vertical asymptote at every integer k

f x csc x

58. limx12

x

2x 1 60. limx1 x 1x4 1 limx1

1x2 1x 1

14

62. limx1

x2 2x 1x 1

64. limx2

13x2 4

52.

(a) Domain:(b)

(c) limx1

f x 0

limx0

f x 0, 0 1,

f x x 1x

68. limx0

cos2 x

x 66. lim

x0

sec x

x


Problem Solving for Chapter 1

2. (a)

(b)

(c) limx0

ax limx0

x 0

ax Area PBOArea PAO x22x2 x

Area PBO 12bh 121y

y2

x2

2

Area PAO 12bh 121x

x

2

x 4 2 1 0.1 0.01

Area 2 1

Area 8 2

4 2 1 1100110ax

120,000120012PBO

120012012PAO

70.

(a)

(b) Yes, define

.

Now is continuous at x 0.f x

f x tan 2xx ,2,x 0

x 0

limx0

tan 2xx

2

f x tan 2xx

x 0.1 0.01 0.001 0.001 0.01 0.1

2.0271 2.0003 2.0000 2.0000 2.0003 2.0271f x

4. (a)

(b) Tangent line:

(c) Let

(d)

This is the slope of the tangent line at P.

limx3

3 x25 x2 4

6

4 4 34

limx3

3 x3 xx 325 x2 4

limx3

25 x2 16x 325 x2 4

limx3

mx limx3

25 x2 4x 3

25 x2 425 x2 4

mx 25 x2 4

x 3

Q x, y x, 25 x2

y 34x

254

y 4 34x 3Slope 34

Slope 4 03 0 43

6.

Letting simplifies the numerator.

Thus,

Setting you obtain

Thus, and b 6.a 3

b 6.b3 3

3,

limx0

b3 bx 3

.

limx0

3 bx 3x

limx0

bxx3 bx 3

a 3

a bx 3

xa bx 3

a bx 3x

a bx 3

xa bx 3a bx 3

8.

Thus,

a 1, 2

a 2a 1 0

a2 a 2 0

a2 2 a

because limx0

tan xx

1limx0

f x limx0

ax

tan x a

limx0

f x limx0

a2 2 a2 2

Problem Solving for Chapter 1 327

12. (a)

Let

(b)

Let

(c)

Let

Since this is smaller than the escape velocity for earth,the mass is less.

v0 6.99 2.64 misec.

limv0

r 10,600

6.99 v02

r 10,600

v2 v02 6.99

v0 2.17 misec 1.47 misec.

limv0

r 1920

2.17 v02

r 1920

v2 v02 2.17

1920r

v2 v02 2.17

v2 1920

r v0

2 2.17

v0 48 43 feetsec.

limv0

r 192,00048 v02

r 192,000

v v02 48

192,000r

v2 v02 48

v2 192,000

r v0

2 48

10.

11

1

2

3

2

1

y

x

(a)

f 1 1 1 f 3 13 0

f 14 4 4 (b)

limx0

f x limx0

f x limx1

f x 0 limx1

f x 1 (c) f is continuous for all real numbers except

x 0, 1, 12, 13, . . .

14. Let and let be given. There exists such that if then Let Then for you have

As a counterexample, let

Then

but limx0

f ax limx0

f 0 2.limx0

f x 1 L,

f x 12 x 0 x 0. f ax L < .

ax < 1

x < 1a

0 < x 0 < 1a, 1a. f x L < .0 < x 0 < ,

1 > 0 > 0a 0



Section 1.1 A Preview of Calculus . . . . . . . . . . . . . . . . . . . . 27

Section 1.2 Finding Limits Graphically and Numerically . . . . . . . . 27

Section 1.3 Evaluating Limits Analytically . . . . . . . . . . . . . . . 31

Section 1.4 Continuity and One-Sided Limits . . . . . . . . . . . . . . 37

Section 1.5 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . 42

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

27


Section 1.1 A Preview of CalculusSolutions to Odd-Numbered Exercises

1. Precalculus: 20 ftsec15 seconds 300 feet 3. Calculus required: slope of tangent line at is rate ofchange, and equals about 0.16.

x 2

5. Precalculus: sq. unitsArea 12 bh 12 53

152 7. Precalculus: cubic unitsVolume 243 24

9. (a)

(b) The graphs of are approximations to the tangent line to at (c) The slope is approximately 2. For a better approximation make the list numbers smaller:

0.2, 0.1, 0.01, 0.001

x 1.y1y2

8

2

4

6

(1, 3)

11. (a)(b)

(c) Increase the number of line segments. 2.693 1.302 1.083 1.031 6.11

D2 1 522 1 52 532 1 53 542 1 54 12D1 5 12 1 52 16 16 5.66

Section 1.2 Finding Limits Graphically and Numerically

1.


x 2x2 x 2 0.3333

x 1.9 1.99 1.999 2.001 2.01 2.1

0.3448 0.3344 0.3334 0.3332 0.3322 0.3226f x

3.

Actual limit is 123.limx0

x 3 3x

0.2887

x 0.1 0.01 0.001 0.001 0.01 0.1

0.2911 0.2889 0.2887 0.2887 0.2884 0.2863f x

5.


1x 1 14x 3 0.0625

x 2.9 2.99 2.999 3.001 3.01 3.1

0.06100.06230.06250.06250.06270.0641f x

7.

(Actual limit is 1.) (Make sure you use radian mode.)limx0

sin xx

1.0000

x 0.1 0.01 0.001 0.001 0.01 0.1

0.9983 0.99998 1.0000 1.0000 0.99998 0.9983f x

9. limx3

4 x 1 11. limx2

f x limx2

4 x 2

15. tan x does not exist since the function increases and

decreases without bound as x approaches 2.

limx2

17. does not exist since the function oscillates

between and 1 as x approaches 0.1

limx0

cos1x

(b)

(c)

does not exist. The values of C jump from 1.75 to 2.25 at t 3.limt3

Ct

limt3.5

Ct 2.25

t 2 2.5 2.9 3 3.1 3.5 4

C 1.25 1.75 1.75 1.75 2.25 2.25 2.25

19.

(a)

00

5

3

Ct 0.75 0.50t 1 t 3 3.3 3.4 3.5 3.6 3.7 4

C 1.75 2.25 2.25 2.25 2.25 2.25 2.25

21. You need to find such that implies

That is,

19 > x 1 >

111.

109 1 > x 1 >

1011 1

109 > x >

1011

910 <

1x

0

0 <

3 3 <

> 0:

limx6

3 3 33.

Given

Hence, let

Hence for you have

f x L < 3x 0 <

3x < x < 3

0 < x 0 < 3, 3.

x < 3

3x <

3x 0 < > 0:

limx0

3x 0

Section 1.2 Finding Limits Graphically and Numerically 29

35.

Given

Hence,

Hence for you have

(because ) f x L <

x 20

x 2 4 < x 2 4 <

x 2 < x 2 <

0 < x 2 < , .

x 2 x 2 x 2 < x 2 < 0 x 2 4 <

x 2 4 < > 0:

limx2

x 2 2 2 4 37.Given

If we assume then

Hence for you have

f x 2 < x2 1 2 <

x2 1 < x 1 < 13 0:

limx1

x2 1 2

39.

The domain is The graphing utility does not show the hole at 4, 16.

5, 4 4, .

limx4

f x 16 6 60.1667

0.5

f x x 5 3x 4 41.

The domain is all except The graphingutility does not show the hole at 9, 6.

x 9.x 0

limx9

f x 6

00

10

10f x x 9x 3

43. means that the values of f approach 25 as x gets closer and closer to 8.limx8

f x 25

45. (i) The values of f approach differentnumbers as x approaches c fromdifferent sides of c:

x1

1

34

2

1

34

2 134 2 3 4

y

47.

x2 3 4 5

2

3

7

1123

1

1

(0, 2.7183)

y

limx0

1 x1x e 2.71828

f x 1 x1x x2.8679722.7319992.7196422.7184182.7182952.7182830.000001

0.000010.00010.0010.010.1

f x x0.1 2.5937420.01 2.7048140.001 2.7169420.0001 2.7181460.00001 2.7182680.000001 2.718280

f x

(ii) The values of f increase with-out bound as x approaches c:

3 2 1

21

1234

56

2 3 4 5x

y

(iii) The values of f oscillatebetween two fixed numbers asx approaches c:

x

34

34

4 3 2 2 3 4

y


Section 1.3 Evaluating Limits Analytically

49. False; isundefined when From Exercise 7, we have

limx0

sin xx

1.

x 0.f x sin xx 51. False; let

limx4

f x limx4

x2 4x 0 10

f 4 10

f x x2 4x,10, x 4x 4.53. Answers will vary.

55. If and then for every there exists and such that and

Let equal the smaller of and Then for we have

Therefore, Since is arbitrary, it follows that L1 L2. > 0L1 L2 < 2.L1 L2 L1 f x f x L2 L1 f x f x L2 < .

x c < ,2.1x c < 2 f x L2 < . f x L1 < x c < 1 2 > 01 > 0 > 0,L2,lim

xc f x L1lim

xc f x

57. means that for every there exists such that if

then

This means the same as when

Thus, limxc

f x L.0 < x c < .

f x L < f x L 0 < .

0 < x c < , > 0 > 0lim

xc f x L 0

1. (a)

(b)

hx x2 5x

limx1

hx 6

limx5

hx 0

13

7

8

7 3. (a)(b)

f x x cos x

6

lim

x3 f x 0.524

limx0

f x 0

4

4

5. limx2

x4 24 16 7. limx0

2x 1 20 1 1

9. limx3

x2 3x 32 33 9 9 0

11. limx3

2x2 4x 1 232 43 1 18 12 1 7

13. limx2

1x

12 15. limx1

x 3x2 4

1 312 4

25

25

17. limx7

5xx 2

57

7 2

359

353

19. limx3

x 1 3 1 2


21. limx4

x 32 4 32 1 23. (a)(b)(c) lim

x1 g f x g f 1 g4 64

limx4

gx 43 64

limx1

f x 5 1 4

25. (a)(b)(c) lim

x1 g f x g3 2

limx3

gx 3 1 2

limx1

f x 4 1 3 27. limx2

sin x sin 2 1

29. limx2

cos x

3 cos 23

12

31. limx0

sec 2x sec 0 1

33. limx56

sin x sin 56 12 35. limx3 tan

x

4 tan 34 1

37. (a)

(b)

(c)

(d) limxc

f xgx

limxc

f xlimxc

gx 23

limxc

f xgx limxc f xlimxc gx 23 6

limxc

f x gx limxc

f x limxc

gx 2 3 5

limxc

5gx 5 limxc

gx 53 15 39. (a)

(b)

(c)

(d) limxc

f x32 limxc

f x32 432 8

limxc

3 f x 3 limxc

f x 34 12

limxc

f x limxc f x 4 2

limxc

f x3 limxc

f x3 43 64


(a)

(b) limx1

gx limx1

f x 3

limx0

gx limx0

f x 1x 0.

gx 2x2 x

xf x 2x 1 43. and agree except at

(a)

(b) limx1

gx limx1

f x 0

limx1

gx limx1

f x 2

x 1.gx x3 x

x 1f x xx 1


3

4

3

4

limx1

f x limx1

gx 2

x 1.gx x 1f x x2 1

x 1 47. and agree except at

90

9

12

limx2

f x limx2

gx 12

x 2.

gx x2 2x 4f x x3 8

x 2

49.

limx5

1x 5

110

limx5

x 5x2 25 limx5

x 5x 5x 5 51.

limx3

x 2x 3

56

56

limx3

x2 x 6x2 9 limx3

x 3x 2x 3x 3


53.

lim x0

x 5 5xx 5 5 limx0

1x 5 5

1

25

510

limx0

x 5 5x

limx0

x 5 5x

x 5 5x 5 5

55.

limx4

x 5 9x 4x 5 3 limx4

1x 5 3

1

9 3

16

limx4

x 5 3x 4 limx4

x 5 3x 4

x 5 3x 5 3

57. limx0

12 x

12

x lim

x0

2 2 x22 x

x lim

x0

122 x

14

59. lim

x0

2x x 2x

x

lim

x0

2x 2x 2x

x lim

x0 2 2

61.

lim

x0

2x x 2 2x 2

lim

x0

x x2 2x x 1 x2 2x 1

x

lim

x0

x2 2xx x2 2x 2x 1 x2 2x 1

x

63.

Analytically,

limx0

x 2 2xx 2 2 limx0

1x 2 2

1

22

24 0.354

limx0

x 2 2x

limx0

x 2 2x

x 2 2x 2 2

2

3 3

2

limx0

x 2 2x

0.354

x 0.1 0.01 0.001 0 0.001 0.01 0.1

0.358 0.354 0.345 ? 0.354 0.353 0.349f x

65.


12 x

12

x lim

x0

2 2 x22 x

1x

limx0

x

22 x 1x

limx0

122 x

14

5 1

2

3

limx0

12 x

12

x

14

x 0.1 0.01 0.001 0 0.001 0.01 0.1

0.263 0.251 0.250 ? 0.250 0.249 0.238f x


67. 115 15limx0

sin x5x limx0

sin xx

15 69.

1210 0

limx0

sin x1 cos x2x2 limx0

12

sin xx

1 cos x

x

71. 1 sin 0 0limx0

sin2 xx

limx0

sin xx sin x 73. 00 0

limh0

1 cos h2h limh0

1 cos hh 1 cos h

75. limx2

cos x

cot x lim

x2 sin x 1 77. lim

t0

sin 3t2t limt0

sin 3t3t

32 1

32

32

79.

Analytically, .limt0

sin 3tt

limt0

3sin 3t3t 31 3

1

4

22

f t sin 3tt

t 0.1 0.01 0.001 0 0.001 0.01 0.1

2.96 2.9996 3 ? 3 2.9996 2.96f t


81.


sin x2x

limx0

xsin x2

x2 01 0

2 2

1

1

f x sin x2

x

x 0 0.001 0.01 0.1

? 0.001 0.01 0.0999980.0010.010.099998f x0.0010.010.1

83. limh0

2hh 2 limh0

2x 2h 3 2x 3hlimh0

f x h f xh limh0

2x h 3 2x 3h

85. limh0

4x hx

4x2

limh0

4x 4x hx hxhlimh0

f x h f xh limh0

4x h

4x

h

87.

Therefore, limx0

f x 4. 4 lim

x0 f x 4

limx0

4 x2 limx0

f x limx0

4 x2 89.

limx0

x cos x 0

4

4

32

32

f x x cos x


91.

limx0

x sin x 0

2 2

6

6

f x x sin x 93.

limx0

x sin 1x 0

0.5

0.5

0.5

0.5

f x x sin 1x

95. We say that two functions f and g agree at all but onepoint (on an open interval) if for all x in theinterval except for where c is in the interval.x c,

f x gx97. An indeterminant form is obtained when evaluating a limit

using direct substitution produces a meaningless fractionalexpression such as That is,

for which limxc

f x limxc

gx 0

limxc

f xgx

00.

99.

3

3

5 5

fg h

hx sin xx

gx sin x,f x x,

When you are close to 0 the magnitude of f isapproximately equal to the magnitude of g.Thus, when x is close to 0.g f 1

101.

Speed 160 ftsec

limt5

s5 st5 t limt5

600 16t2 10005 t limt5

16t 5t 5t 5 limt5 16t 5 160 ftsec.

st 16t2 1000

103.

limx3

4.93 t3 t3 t limx3 4.93 t 29.4 msec

limt3

s3 st3 t limt3

4.932 150 4.9t2 1503 t limt3

4.99 t23 t

st 4.9t2 150

105. Let and and do not exist.

limx0

f x gx limx0

1x 1x limx0 0 0

limx0

gxlimx0

f xgx 1x.f x 1x

107. Given show that for every there exists a such that whenever . Sincefor any then any value of will work. > 0 > 0, f x b b b 0 <

x c < f x b < > 0 > 0f x b,

109. If then the property is true because both sides are equal to 0. If let be given. Since

there exists such that whenever . Hence, wherever

we have

or

which implies that limxc

bf x bL.bf x bL < b f x L <

0 < x c < ,0 < x c < f x L < b > 0limxc

f x L, > 0b 0,b 0,


111.

Therefore, limx c

f xgx 0. 0 lim

xc f xgx 0

M0 limxc

f xgx M0limxc

M f x limxc

f xgx limxc

M f x M f x f xgx M f x 113. False. As x approaches 0 from the left,

3

2

3

2

xx

1.

115. True. 117. False. The limit does not exist.

6

2

3

4

119. Let

does not exist since for and for f x 4.x 0,f x 4x < 0,limx0

f x

limx0

f x limx0

4 4.

f x 4,4, if x 0if x < 0

121.

does not exist.

No matter how close to 0 x is, there are still an infinite number of rational and irrational numbers so that does notexist.

When x is close to 0, both parts of the function are close to 0.

limx0

gx 0.

f xlimx0

limx0

f x

g x 0,x, if x is rational if x is irrational

f x 0,1, if x is rational if x is irrational

123. (a)

112 12

limx0

sin2 xx2

1

1 cos x

limx0

1 cos2 xx21 cos x

limx0

1 cos xx2

limx0

1 cos xx2

1 cos x1 cos x (b) Thus,

for

(c)

(d) which agrees with part (c).cos0.1 0.9950,

cos0.1 1 120.12 0.995

x 0. cos x 1 12x2

1 cos xx2

12 1 cos x

12x

2


Section 1.4 Continuity and One-Sided Limits

1. (a)

(b)

(c)

The function is continuous atx 3.

limx3

f x 1

limx3

f x 1

limx3

f x 1 3. (a)

(b)

(c)


limx3

f x 0

limx3

f x 0

limx3

f x 0 5. (a)(b)(c) does not existThe function is NOT continuousat x 4.

limx4

f xlim

x4 f x 2

limx4

f x 2

7. limx5

x 5x2 25 limx5

1x 5

110 9. does not exist because grows

without bound as x3.

x

x2 9lim

x3

x

x2 9

11. limx0

xx

limx0

x

x 1.

13.

1x2

1

xx 0

limx0

1xx x

limx0

1x x

1x

x lim

x0

x x xxx x

1x

limx0

x

xx x 1

x

15. limx3

f x limx3

x 22

52 17.

limx1

f x 2

limx1

f x limx1

x3 1 2

limx1

f x limx1

x 1 2

19. does not exist since

and do not exist.limx

cot xlimx

cot x

limx

cot x 21.x 3 for 3 < x < 4lim

x4 3x 5 33 5 4

23. does not exist

because

and

limx3

2 x 2 4 6.

limx3

2 x 2 3 5

limx3

2 x 25.

has discontinuities at andsince and are not

defined.f 2f 2x 2

x 2

f x 1x2 4 27.

has discontinuities at each integerk since lim

xk f x lim

xk f x.

f x x2 x

29. is continuouson 5, 5.gx 25 x2 31.

f is continuous on 1, 4.lim

x0 f x 3 lim

x0 f x. 33. is continuous

for all real x.f x x2 2x 1


35. is continuous for all real x.f x 3x cos x 37. is not continuous at Since

for is a removable

discontinuity, whereas is a nonremovablediscontinuity.

x 1

x 0, x 0xx2 x

1

x 1

x 0, 1.f x xx2 x

39. is continuous for all real x.f x xx2 1 41.

has a nonremovable discontinuity at since does not exist, and has a removable discontinuity at

since

limx2

f x limx2

1x 5

17.

x 2

limx5

f xx 5

f x x 2x 2x 5

43. has a nonremovable discontinuity at since does not exist.limx2

f xx 2f x x 2x 2

45.

has a possible discontinuity at .

1.

2.

3.

f is continuous at therefore, f is continuous for all real x.x 1,

f 1 limx1

f x

limx1

f x 1limx1 f x limx1 x 1lim

x1f x lim

x1x2 1

f 1 1x 1

f x x,x2, x 1x > 1

49. has possible discontinuities at

1.

2.

3.

f is continuous at therefore, is continuous for all real x.fx 1,

f 1 limx1

f xf 1 limx1

f x

limx1

f x 1limx1

f x 1f 1 1f 1 1

x 1.x 1, tan x4

,

x,

1 < x < 1x 1 or x 1f x tan

x4 ,

x,

x < 1x 1

47. has a possible discontinuity at

1.

2.

Therefore, f has a nonremovable discontinuity at x 2.

limx2

f x limx2

3 x 1

limx2

f x limx2

x2 1 2f 2 22 1 2

x 2.f x x

2 1,3 x,

x 2

x > 2

does not exist.limx2 f x


51. has nonremovable discontinuities at integermultiples of 2.f x csc 2x 53. has nonremovable discontinuities at each

integer k.f x x 1

55.

f is not continuous at x 2.

limx0

f x 08 8

10

50limx0

f x 0 57.

Find a so that a 822 2.limx2 ax2 8

f 2 8

59. Find a and b such that and

b 2 1 1

a 1

4a 4

3a b 2

a b 2

limx3

ax b 3a b 2.limx1

ax b a b 2

f x 2,x 1,2,

x 11 < x < 3x 3

61.

Continuous for all real x.

f gx x 12 63.

Nonremovable discontinuities at x 1

f gx 1x2 5 6 1

x2 1

65.

Nonremovable discontinuity at each integer

3 3

0.5

1.5

y x x 67.

Nonremovable discontinuity at

5

5

5

7

x 3

f x 2x 4,x2 2x, x 3x > 3

69.

Continuous on ,

f x xx2 1 71.

Continuous on:. . . , 6, 2, 2, 2, 2, 6, 6, 10, . . .

f x sec x4

73.

The graph appears to be continuous on the intervalSince is not defined, we know that f has

a discontinuity at This discontinuity is removableso it does not show up on the graph.

x 0.f 04, 4.

4 4

2

3

f x sin xx


and By the Intermediate ValueTheorem, for at least one value of c between1 and 2.

f c 0f 2 4.f 1 3316

1, 2.f x 116x4 x3 3



and By the IntermediateValue Theorem, for the least one value of cbetween 0 and .

f c 0f 2 1 > 0.f 0 3

0, .f x x2 2 cos x 79.is continuous on

and

By the Intermediate Value Theorem, for at leastone value of c between 0 and 1. Using a graphing utility,we find that x 0.6823.

f x 0f 1 1f 0 1

0, 1.f xf x x3 x 1

81.

is continuous on

and

By the Intermediate Value Theorem, for at leastone value c between 0 and 1. Using a graphing utility, wefind that t 0.5636.

gt 0

g1 1.9 < 0.g0 2 > 0

0, 1.g

gt 2 cos t 3t 83.




x 4c 3

x 4 or x 3

x 4x 3 0

x2 x 12 0

x2 x 1 11

1 < 11 < 29

f 5 29f 0 10, 5.

f x x2 x 1

85.



( has no real solution.)

Thus, f 2 4. c 2

x2 x 3

x 2

x 2x2 x 3 0

x3 x2 x 6 0

x3 x2 x 2 4

2 < 4 < 19

f 3 19f 0 20, 3.

f x x3 x2 x 2 87. (a) The limit does not exist at (b) The function is not defined at (c) The limit exists at but it is not equal to the

value of the function at

(d) The limit does not exist at x c.

x c.x c,

x c.

x c.

89.

The function is not continuous at because.f x 1 0 lim

x3 f xlim

x3

x 3

x

12 1 3 4 5 6 7

23

12345

y 91. The functions agree for integer values of x:

However, for non-integer values of x, the functionsdiffer by 1.

For example, f 12 3 0 3, g12 3 1 4.f x 3 x gx 1 2 x.

gx 3 x 3 x 3 xf x 3 x 3 x for x an integer


t2 4 6 8 10 12

10

20

30

40

50

N

Time (in months)

Num

ber o

f uni

ts

93.

Discontinuous at every positive even integer.The company replenishes its inventory every two months.

Nt 252t 22 t

95. Let be the volume of a sphere of radius r.

Since 4.19 < 275 < 523.6, the Intermediate Value Theorem implies that there is at least one value r between 1 and 5 suchthat (In fact, .)r 4.0341Vr 275.

V5 43 53 523.6

V1 43 4.19

V 43 r3

97. Let c be any real number. Then does not exist since there are both rational and

irrational numbers arbitrarily close to c. Therefore, f is not continuous at c.limxc

f x

99.

(a)

(b)

(c) does not exist.limx0

sgnx

limx0

sgnx 1

limx0

sgnx 1

sgnx 1,0,1,if x < 0if x 0if x > 0

x1

2

34

2

1

34

2 134 2 3 4

y

t 0 1 1.8 2 3 3.8

50 25 5 50 25 5Nt

101. True; if then andat least one of these limits (if they exist) does not equalthe corresponding function at x c.

limxc

f x limxc

gxx c,f x gx, 103. False; is not defined and does not exist.limx1

f xf 1

105. (a)

NOT continuous at x b.

x2b

2b

b

b

y

0 b 0 x < bb < x 2bf x (b)

Continuous on 0, 2b.

x2b

2b

b

b

y

x

2

b x 2

0 x b

b < x 2bgx


Section 1.5 Infinite Limits

107.

Domain: and

Define to make f continuous at x 0.f 0 12c

limx0

x c2 c2

xx c2 c limx0 1

x c2 c

12c

limx0

x c2 cx

limx0

x c2 cx

x c2 c

x c2 c

c2, 0 0, x 0,x c2 0 x c2

f x x c2 c

x, c > 0

109.

h has nonremovable discontinuities at x 1, 2, 3, . . . .

3

3

3

15hx xx

1.

limx2

2 xx2 4 lim

x2 2 xx2 4 3.

limx2

tan x

4

limx2

tan x

4

5.

limx3

f x

limx3

f x

f x 1x2 9

x

0.308 1.639 16.64 0.3641.69516.69166.7166.6f x2.52.92.992.9993.0013.013.13.5

7.

limx3

f x

limx3

f x

f x x2

x2 9

x

3.769 15.75 150.8 2.27314.25149.314991501f x2.52.92.992.9993.0013.013.13.5


9.


limx0

1x2

limx0

1x2

11.



limx1

x2 2x 2x 1

limx1

x2 2x 2x 1

x 2

limx2

x2 2x 2x 1

limx2

x2 2x 2x 1

13. and


and


limx2

x2

x2 4 limx2 x2

x2 4

x 2

limx2

x2

x2 4 limx2 x2

x2 4 15. No vertical asymptote since the denominator is never zero.


n any integer.x 2n 14

4 n

2 ,

f x tan 2x sin 2xcos 2x 19.

Therefore, is a vertical asymptote.t 0

limt0

1 4t 2 limt0 1 4t 2

21.



limx1

x

x 2x 1

limx1

x

x 2x 1

x 2

limx2

x

x 2x 1

limx2

x

x 2x 1 23.

has no vertical asymptote since

limx1

f x limx1

x2 x 1 3

f x x3 1

x 1 x 1x2 x 1

x 1

25.

No vertical asymptotes. The graph has a hole at x 5.

f x x 5x 3x 5x2 1 x 3x2 1, x 5 27. has vertical asymptotes at n

a nonzero integer. There is no vertical asymptote at since

limt0

tsin t 1.

t 0

t n,st tsin t


29.

Removable discontinuity at x 1

3 3

5

2

limx1

x2 1x 1 limx1 x 1 2 31.

Vertical asymptote atx 1

limx1

x2 1x 1

3 3

8

8limx1

x2 1x 1

33. limx2

x 3x 2 35. limx3

x2

x 3x 3

37. limx3

x2 2x 3x2 x 6 limx3

x 1x 2

45 39. limx1

x2 x

x2 1x 1 limx1 x

x2 1 12

41. limx0

1 1x 43. limx0 2

sin x

45. limx

xcsc x

limx

x sin x 0 47. and Therefore, does not exist.lim

x12 x secx

limx12

x secx .limx12

x secx

49.

5

3

4

3

limx1

f x limx1

1x 1

f x x2 x 1x3 1 51.

8 8

0.3

0.3

limx5

f x

f x 1x2 25

53. A limit in which increases or decreases withoutbound as x approaches c is called an infinite limit. isnot a number. Rather, the symbol

says how the limit fails to exist.

limxc

f x

f x 55. One answer is f x x 3x 6x 2

x 3x2 4x 12.

57. y

x1 3 1 2

1

2

2

1

3

59. Assume

(or if )k < 0limr1

S limr1

k1 r

k 0.0 < r < 1.S k1 r,


(d)

For n 3, limx0

x sin xxn

.

limx0

x sin xx4

1.5

1.5

1.5

1.5

61.

(a) million(b) million(c) million

(d) Thus, it is not possible.limx100

528100 x

C75 $1584C50 $528C25 $176

0 x < 100C 528x100 x, 63. (a)

(b)

(c) limx25

2x625 x2

r 215

625 225

32 ftsec

r 27

625 49

712 ftsec

x 1

1667.0166.716.671.66580.83170.32920.1585f x0.00010.0010.010.10.20.5

65. (a)

limx0

x sin xx

0

1.5

0.25

1.5

0.5

x 1

0000.00170.00670.04110.1585f x0.00010.0010.010.10.20.5

(b)

limx0

x sin xx2

01.5

0.25

1.5

0.25

x 1

000.00170.01670.03330.08230.1585f x0.00010.0010.010.10.20.5

(c)

limx0

x sin xx3

0.1167 161.5

0.25

1.5

0.25

x 1

0.16670.16670.16670.16660.16630.16460.1585f x0.00010.0010.010.10.20.5


67. (a) Because the circumference of the motor is half that of the saw arbor, the saw makes

revolutions per minute.

(c) straight sections.The angle subtended in each circle is

Thus, the length of the belt around the pulleys is

Total length

Domain: 0, 2 60 cot 30 2

20 2 10 2 30 2.

2 22 2.

220 cot 210 cot :

17002 850

(b) The direction of rotation is reversed.(d)

(e)

(f)

(All the belts are around pulleys.)(g) lim

0 L

lim2

L 60 188.5

00

450

2

0.3 0.6 0.9 1.2 1.5

L 306.2 217.9 195.9 189.6 188.5

69. False; for instance, let

or

.gx xx2 1

f x x2 1

x 1

71. False; let

The graph of f has a vertical asymptote at butf 0 3.

x 0,

f x 1x

,

3,x 0x 0.

73. Given and

(2) Product:If L > 0, then for there exists such that whenever Thus,

Since then for M > 0, there exists such that wheneverLet be the smaller of and Then for we have

Therefore The proof is similar for L < 0.

(3) Quotient: Let be given.There exists such that whenever and there exists such that

whenever This inequality gives us Let be the smaller of and Thenfor we have

Therefore, limxc

gxf x 0.

gxf x < 3L23L2 .0 < x c < ,

2.1L2 < gx < 3L2.0 < x c < 2.L2

gx L 00 < x c < 1f x > 3L21 > 0

> 0

limxc

f xgx .f xgx > M2LL2 M.0 < x c < ,2.1x c < 2.

f x > M2L2 > 0limxc

f x L2 < gx < 3L2.0 < x c < 1.gx L < L21 > 0 L2 > 0

limxc

gx L:limxc

f x

75. Given

Suppose exists and equals L. Then,

This is not possible. Thus, does not exist.limxc

f x

limxc

1f x

limxc

1

limxc

f x 1L 0.

limxc

f x

limxc

1f x 0.


Review Exercises for Chapter 1

1. Calculus required. Using a graphing utility, you can estimate the length to be 8.3.Or, the length is slightly longer than the distance between the two points, 8.25.

3.

limx0

f x 0.25

1 1

1

1x 0.001 0.01 0.1

0.240.2490.24990.2500.250.26f x0.0010.010.1

5. (a)

(b) limx1

hx 3

limx0

hx 2hx x2 2x

x7.

Let be given. Choose Then for

you have

f x L < 3 x 2 <

1 x < x 1 <

0 < x 1 < , . > 0

limx1

3 x 3 1 2

9.

Let be given. We need

Assuming, you can choose Hence, for you have

f x L < x2 3 1 <

x2 4 < x 2x 2 <

x 2 < 5 0limx2

x2 3 1

11. limt4

t 2 4 2 6 2.45 13. limt2

t 2t2 4 limt2

1t 2

14

15.

limx4

1x 2

1

4 2

14

limx4

x 2x 4

limx4

x 2x 2x 2 17.

limx0

1x 1 1

limx0

1x 1 1x

limx0

1 x 1xx 1

19.

75

limx5

x2 5x 25

limx5

x3 125x 5 limx5

x 5x2 5x 25x 5 21. limx0

1 cos xsin x limx0

x

sin x

1 cos x

x 10 0

Review Exercises for Chapter 1 47

23.

0 32 1

32

limx0

12

cos x 1x

limx0

32

sin xx

limx0

sin6 x 12x

limx0

sin6 cos x cos6 sin x 12x

25. limxc

f x gx 3423 12

27.

(a)

Actual limit is

(c)

2

23

13

33

limx1

22x 1 3

limx1

2x 1 3x 12x 1 3

limx1

2x 1 3x 1 limx1

2x 1 3x 1

2x 1 32x 1 3

33.limx1

2x 1 3x 1 0.577

f x 2x 1 3x 1

(b)

20

1

2x 1.1 1.01 1.001 1.0001

0.5680 0.5764 0.5773 0.5773f x

29.

limt4

4.9t 4 39.2 msec

limt4

4.9t 4t 44 t

limta

sa sta t

limt4

4.942 200 4.9t2 2004 t 31. limx3

x 3x 3 limx3

x 3x 3 1

33. limx2

f x 0 35. does not exist because andlim

t1 ht 121 1 1.

limt1

ht 1 1 2limt1

ht

37.where k is an integer.

where k is an integer.

Nonremovable discontinuity at each integer kContinuous on for all integers kk, k 1

limxk

x 3 k 2

limxk

x 3 k 3f x x 3 39.


x 1

limx1

f x limx1

3x 2 5

f x 3x2 x 2x 1

3x 2x 1x 1

41.


x 2

limx2

1x 22

f x 1x 22 43.


x 1

limx1

f x

lim

x1 f x

f x 3x 1


69.

(a) $14,117.65 (b) $80.000

(c) $720,000 (d) limp100

80,000p100 p C90

C50 C15

C 80,000p100 p, 0 0 < 100

45.

Nonremovable discontinuities at each even integer.Continuous on

for all integers k.

2k, 2k 2

f x csc x2 47.Find c so that

c 12

2c 1

c2 6 5

limx2

cx 6 5.

f 2 5

49. f is continuous on andTherefore by the Intermediate Value

Theorem, there is at least one value c in suchthat 2c3 3 0.

1, 2f 2 13 > 0.

f 1 1 < 01, 2.

53.

Vertical asymptote at x 0

gx 1 2x

55.

Vertical asymptote at x 10

f x 8x 102

57. limx2

2x2 x 1x 2

59. limx1

x 1x3 1 limx1

1x2 x 1

13

61. limx1

x2 2x 1x 1 63. limx0 x

1x3

65. limx0

sin 4x5x limx0

45

sin 4x4x

45 67. limx0

csc 2xx

limx 0

1x sin 2x

51.

(a)

(b)

(c) does not exist.limx2

f x

limx2

f x 4

limx2

f x 4

f x x2 4

x 2 x 2x 2

x 2

Problem Solving for Chapter 1

1. (a)

(c) limx0

rx 1 0 11 0 1 22 1

x 12 x4 x2 x4 1

Perimeter PBO x 12 y2 x2 y2 1

x2 x2 12 x2 x4 1

Perimeter PAO x2 y 12 x2 y2 1

x 4 2 1 0.1 0.01

Perimeter 33.02 9.08 3.41 2.10 2.01

Perimeter 33.77 9.60 3.41 2.00 2.00

0.98 0.95 1 1.05 1.005rx

PBO

PAO

(b) rx x2 x2 12 x2 x4 1

x 12 x4 x2 x4 1


5. (a)

(b) Slope of tangent line is

Tangent line

(c)

(d)

This is the same slope as part (b).

10

12 12 5

12

limx5

x 512 169 x2

limx5

x2 25x 512 169 x2

limx5

144 169 x2x 512 169 x2

limx5

mx limx5

12 169 x2x 5

12 169 x212 169 x2

mx 169 x2 12

x 5

Q x, y x, 169 x2

y 5

12x 16912

y 12 512x 5

512.

Slope 1253. (a) There are 6 triangles, each with a central angle of

Hence,

Error:

(b) There are n triangles, each with central angle ofHence,

(c)

(d) As n gets larger and larger, approaches 0.Letting

which approaches 1 .

An sin2n2n sin2n

2n sin x

x

x 2n,

2n

An n12bh n121 sin

2n

n sin 2n2 .

2n.

33

2 0.5435.

1

h = sin

60

1

h = sin 60

33

2 2.598.

Area hexagon 612bh 6121 sin

3

60 3.

n 6 12 24 48 96

An 2.598 3 3.106 3.133 3.139

(d)

1

1 1 12 2 1

12

limx1

1x23 x13 13 x13 2

limx1

x13 1x13 1x23 x13 13 x13 2

limx1

3 x13 4x 13 x13 2

limx1

f x limx1

3 x13 2x 1

3 x13 23 x13 2

7. (a)

Domain: x 27, x 1

x 27

x13 3

3 x13 0 (b)

12

0.1

30

0.5 (c)

2

28 1

14 0.0714

limx27

f x 3 2713 2

27 1

9. (a)(b) f continuous at 2:(c) lim

x2 f x 3: g1, g3, g4

g1

limx2

f x 3: g1, g4


11.

(a)

(b)

(c) f is continuous for all real numbers exceptx 0, 1, 2, 3, . . .

limx12

f x 1 limx1

f x 1 limx1

f x 1 f 2.7 3 2 1

f 12 0 1 1 f 0 0 f 1 1 1 1 1 0

x1

2

34

2

1

34

2 134 2 3 4

y 13. (a)

(b) (i)(ii)(iii)(iv)

(c) is continuous for all positive real numbersexcept

(d) The area under the graph of u,and above the x-axis, is 1.

x a, b.Pa, b

limxb

Pa, bx 1

limxb

Pa, bx 0

limxa

Pa, bx 0

limxa

Pa, bx 1

xb

2

a

1

y


C H A P T E R 2Differentiation

Section 2.1 The Derivative and the Tangent Line Problem . . . 53

Section 2.2 Basic Differentiation Rules and Rates of Change . 60

Section 2.3 The Product and Quotient Rules andHigher-Order Derivatives . . . . . . . . . . . . . . 67

Section 2.4 The Chain Rule . . . . . . . . . . . . . . . . . . . 73

Section 2.5 Implicit Differentiation . . . . . . . . . . . . . . . 79

Section 2.6 Related Rates . . . . . . . . . . . . . . . . . . . . 85

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 92

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . 98

53

C H A P T E R 2Differentiation

Section 2.1 The Derivative and the Tangent Line ProblemSolutions to Odd-Numbered Exercises

1. (a)(b) m 3

m 0 3. (a), (b) (c)

x 1

1x 1 2

33 x 1 2

6

5

4

3

2

654321

1

y

x

1)1f ) x

)1 3f ))f )4

11)f )

1 )x))f 4

4y

4) , )5

2)21) ,

f ))1

f ) )4 5

)

y f 4 f 1

4 1 x 1 f 1)

5. is a line. Slope 2f x 3 2x 7. Slope at

limx0

2 2x 2

limx0

1 2x x2 1x

limx0

1 x2 4 3x

1, 3 limx0

g1 x g1x

9. Slope at

limt0

3 t 3

limt0

3t t2 0t

0, 0 limt0

f 0 t f 0t

11.

limx0

0 0

limx0

3 3x

f x limx0

f x x f xx

f x 3

13.

limx0

5 5

limx0

5x x 5xx

fx limx0

f x x f xx

f x 5x 15.

lims0

23s

s

23

lims0

3 23s s 3 23s

s

hs lims0

hs s hss

hs 3 23 s

54 Chapter 2 Differentiation

17.

limx0

4xx 2x2 xx

limx0

4x 2x 1 4x 1

limx0

2x2 4xx 2x2 x x 1 2x2 x 1x

limx0

2x x2 x x 1 2x2 x 1x

fx limx0

f x x f xx

f x 2x2 x 1

19.

limx0

3x2 3xx x2 12 3x2 12

limx0

3x2x 3xx2 x3 12xx

limx0

x3 3x2x 3xx2 x3 12x 12x x3 12xx

limx0

x x3 12x x x3 12xx

fx limx0

f x x f xx

f x x3 12x

21.

1

x 12

limx0

1x x 1x 1

limx0

x

xx x 1x 1

limx0

x 1 x x 1xx x 1x 1

limx0

1x x 1

1x 1

x

fx limx0

f x x f xx

f x 1x 1

23.

1

x 1 x 1

12x 1

limx0

1x x 1 x 1

limx0

x x 1 x 1xx x 1 x 1

limx0

x x 1 x 1x

x x 1 x 1x x 1 x 1

fx limx0

f x x f xx

f x x 1

25. (a)

At the slope of the tangent line isThe equation of the tangent line is

y 4x 3.

y 5 4x 8

y 5 4x 2

m 22 4.2, 5,

limx0

2x x 2x

limx0

2xx x2

x

limx0

x x2 1 x2 1x

fx limx0

f x x f x

x

f x x2 1 17. (b)(2, 5)

5 5

2

8

27. (a)

At the slope of the tangent is The equation of the tangent line is

y 12x 16.

y 8 12x 2

m 322 12.2, 8,

limx0

3x2 3xx x2 3x2

limx0

3x2x 3xx2 x3x

limx0

x x3 x3

x

fx limx0

f x x f xx

f x x3 18. (b)

5 5

4

(2, 8)10

29. (a)

At the slope of the tangent line is

The equation of the tangent line is

y 12 x

12.

y 1 12 x 1

m 1

21 12.

1, 1,

limx0

1x x x

1

2x

limx0

x x xxx x x

limx0

x x xx

x x xx x x

fx limx0

f x x f xx

f x x 18. (b)

5

1

1

3

(1, 1)

Section 2.1 The Derivative and the Tangent Line Problem 55


31. (a)

At the slope of the tangent line is

The equation of the tangent line is

y 34x 2

y 5 34x 4

m 1 416 34

4, 5,

x2 4

x2 1 4

x2

limx0

x2 xx4xx x

limx0

x2x xx2 4xxxx x

limx0

x3 2x2x xx2 x3 x2x 4xxxx x

limx0

xx xx x 4x x2x x 4x xxxx x

limx0

x x 4x x

x 4xx

fx limx0

f x x f xx

f x 4x

(b)

12

6

12

10

(4, 5)

33. From Exercise 27 we know that Since theslope of the given line is 3, we have

Therefore, at the points and the tangentlines are parallel to These lines haveequations

and

y 3x 2. y 3x 2

y 1 3x 1y 1 3x 1

3x y 1 0.1, 11, 1

x 1.

3x2 3

fx 3x2. 35. Using the limit definition of derivative,

Since the slope of the given line is , we have

Therefore, at the point the tangent line is parallel toThe equation of this line is

y 12x

32.

y 1 12x 12

y 1 12x 1

x 2y 6 0.1, 1

x 1.

1

2xx 12

12

fx 12xx.

37. because the tangent line passes through

g5 2 05 9 2

4 12

5, 2g5 2 39. f x x fx 1 b


41. matches (a) decreasing slope as x

f x x fx 43.

Answers will vary.

Sample answer: y x

x1

1

2

34

2

1

34

2 134 2 3 4

y

45. (a) If and is odd, then (b) If and is even, then fc f c 3ffc 3

fc fc 3ffc 3

47. Let be a point of tangency on the graph of f. By the limit definition for the derivative, The slope of theline through and equals the derivative of f at

Therefore, the points of tangency are and and the corresponding slopes are 2 and . The equations of the tangentlines are

y 2x 1 y 2x 9

y 5 2x 2 y 5 2x 2

23, 3,1, 3

0 x0 1x0 3 x0 1, 3

0 x02 4x0 3

5 4x0 x02 8 8x0 2x02 5 y0 2 x04 2x0

5 y02 x0

4 2x0

x1 2 3 62

2

43

5

7

6

1

(1, 3)(3, 3)

(2, 5)

yx0:x0, y02, 5fx 4 2x.x0, y0

49. (a)(b)(c) Because g is decreasing (falling) at (d) Because g is increasing (rising) at (e) Because and are both positive, is greater than , and (f) No, it is not possible. All you can say is that g is decreasing (falling) at x 2.

g6 g4 > 0.g4g6g6g4

x 4.g4 73 ,

x 1.g1 83 ,

g3 0

g0 3

51.

By the limit definition of the derivative we have fx 34 x2.2

2

2

2

f x 14 x3

x 0 0.5 1 1.5 2

0 2

3 0 3271634

316

316

34

2716fx

2732

14

132

132

14

27322f x

0.511.52


53.

The graph of is approximately the graph of fx.gx

3

1

2 4

g

f

2x 0.01 x 0.012 2x x2 100

gx f x 0.01 f x0.0155.

Exact: f2 0f2 3.99 42.1 2 0.1

f 2.1 2.14 2.1 3.99f 2 24 2 4,

57. and

As is nearly horizontal and thus f 0.x, f5

5

2 5

f

f

fx 12x32 .f x 1x

59.

(a)

(b) As the line approaches the tangent line to at 2, 3.fx0,

x 0.1: Sx 1910x 2 3 1910 x

45

x 0.5: Sx 32x 2 3 32 x

5

1

2 7

f

S0.1

S1

S0.5

x 1: Sx x 2 3 x 1

4 2 x 32 3

xx 2 3 1 x 1

2

xx 2 3 x 2x 2 3

Sx x f 2 x f 2

xx 2 f 2

f x 4 x 32

61.

f2 limx2

f x f 2x 2 limx2

x2 1 3x 2 limx 2

x 2x 2x 2 limx2 x 2 4

f x x2 1, c 2

63.

f2 limx2

f x f 2x 2 limx 2

x2x 2x 2 limx2

x3 2x2 1 1x 2 limx2 x

2 4

f x x3 2x2 1, c 2

65.

Does not exist.

As

As x 0, x

x

1x

x 0, x

x

1x

g0 limx0

gx g0x 0 limx0

xx

.

gx x, c 0 67.

Does not exist.

limx6

1x 613

limx6

x 623 0x 6

f6 limx6

f x f 6x 6

f x x 623, c 6


69.

Does not exist.

limx5

x 5x 5

limx5

x 5 0x 5

h5 limx5

hx h5x 5

h x x 5, c 5

73. is differentiable everywhere except at (Discontinuity)

x 1.f x 75. is differentiable everywhere except at (Sharp turn in the graph)

x 3.f x

77. is differentiable on the interval (At the tangent line is vertical)x 1

1, .f x 79. is differentiable everywhere except at (Discontinuity)

x 0.f x

81.

The derivative from the left is

The derivative from the right is

The one-sided limits are not equal. Therefore, f is not differentiable at x 1.

limx1

f x f 1x 1 limx1

x 1 0x 1 1.

limx1

f x f 1x 1 limx1

x 1 0x 1 1.

f x x 1 83.



These one-sided limits are equal. Therefore, f is differentiable at x 1. f1 0

limx1

x 1 0.

limx1

f x f 1x 1 limx1

x 12 0x 1

limx1

x 12 0.

limx1

f x f 1x 1 limx1

x 13 0x 1

f x x 13,x 12, x 1x > 1

85. Note that f is continuous at



The one-sided limits are equal. Therefore, f is differentiable at x 2. f2 4

limx2

f x f 2x 2 limx2

4x 3 5x 2 limx2 4 4.

limx2

f x f 2x 2 limx2

x2 1 5x 2 limx2 x 2 4.

f x x2 1,4x 3, x 2x > 2x 2.

71. is differentiable everywhere except at (Sharp turn in the graph.)

x 3.f x

87. (a) The distance from to the line is

.

(b)

The function d is not differentiable at This corresponds to the linewhich passes through the point 3, 1.y x 4,

m 1.

5

1

4 4

m3 11 4

m2 1 3m 3m2 1

d Ax1 By1 CA2 B2

x

2

3

1 2 3 4

1

ymx y 4 03, 1


Section 2.2 Basic Differentiation Rules and Rates of Change

89. False. the slope is limx0

f 2 x f 2x

.

93.

Using the Squeeze Theorem, we have Thus, and f is continuous atUsing the alternative form of the derivative we have

Since this limit does not exist (it oscillates between and 1), the function is not differentiable at

Using the Squeeze Theorem again we have Thus, and f is continu-ous at Using the alternative form of the derivative again we have

Therefore, g is differentiable at x 0, g0 0.

limx0

f x f 0x 0 limx0

x2 sin1x 0x 0 limx0 x sin

1x

0.

x 0.limx0

x2 sin1x 0 f 0x2 x2 sin1x x2, x 0.

gx x2 sin1x,0, x 0x 0x 0.1

limx0

f x f 0x 0 limx0

x sin1x 0x 0 limx0 sin

1x.

x 0.limx0

x sin1x 0 f 0x x sin1x x, x 0.

f x x sin1x,0, x 0x 0

91. False. If the derivative from the left of a point does not equal the derivative from the right of a point, then the derivative doesnot exist at that point. For example, if then the derivative from the left at is and the derivative from theright at is At the derivative does not exist.x 0,1.x 0

1x 0f x x,

1. (a) (b)

y1 32

y 32 x12 y x32

y1 12

y 12 x12 y x12 (c) (d)

y1 3

y 3x2 y x3

y1 2

y 2x

y x2

3.

y 0

y 8 5.

y 6x5y x6 7.

y 7x8 7x8

y 1x7

x7 9.

y 15x

45 1

5x45

y 5x x15

11.

fx 1 f x x 1 13.

fx 4t 3 f t 2t2 3t 6 15.

gx 2x 12x2gx x2 4x3 17.

st 3t2 2

st t3 2t 4

19.

y

2 cos sin

y

2 sin cos 21.

y 2x 12 sin x

y x2 12 cos x 23.

y 1x2

3 cos x

y 1x

3 sin x

31.

f1 6

fx 6x3 6x3

f x 3x2

3x2, 1, 3 33.

f0 0

fx 215 x2

f x 12 75x

3, 0, 12 35.

y0 4

y 8x 4

4x2 4x 1

y 2x 12, 0, 1

37.

f0 41 1 3f 4 cos 1 f 4 sin , 0, 0 39.

fx 2x 6x3 2x 6x3

f x x2 5 3x2 41.

gt 2t 12t4 2t 12t4

gt t2 4t3

t2 4t3

43.

fx 1 8x3

x3 8

x3

f x x3 3x2 4

x2 x 3 4x2 45.

y 3x2 1

y xx2 1 x3 x

47.

fx 12x12 2x23 1

2x

2x23

f x x 6 3x x12 6x13 49.

h(s 45s45

23s

13 4

5s15 2

3s13

hs s45 s23

51.

fx 3x12 5 sin x 3x

5 sin x

f x 6x 5 cos x 6x12 5 cos x

53. (a)

At .

Tangent line:

(b) 3

1

2 2(1, 0)

2x y 2 0

y 0 2x 1

y 413 61 21, 0:

y 4x3 6x

y x4 3x2 2

Function Rewrite Derivative Simplify25. y 5

x 3y 5x3y 52x

2y 5

2x2

27. y 98x4y 98 x

4y 38x

3y 3

2x3

29. y 12x32y 12x

32y x12y xx

55. (a)

At

Tangent line:

(b)

7

1

2

5

(1, 2)

3x 2y 7 0

y 32x

72

y 2 32x 1

f1 321, 2,

fx 32 x74

32x74

f x 24x3 2x34

Section 2.2 Basic Differentiation Rules and Rates of Change 61


57.

Horizontal tangents: , 2, 142, 140, 2,

y 0 x 0, 2

4xx 2x 2

4xx2 4

y 4x3 16x

y x4 8x2 2 59.

cannot equal zero.

Therefore, there are no horizontal tangents.

y 2x3 2x3

y 1x2

x2

61.

At .

Horizontal tang

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