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Review Exercises for Chapter 1

2. Precalculus.L 9 12 3 12 8.25

4.

limx0

fx 0.2

1

0.5

1

0.5

x 0.1 0.01 0.001 0.001 0.01 0.1

0.358 0.354 0.354 0.354 0.353 0.349fx

6. (a) does not exist.limx2

gxgx 3x

x 2(b) lim

x0gx 0

8.

Let be given. We need

x 9 < x 3

x

3<

x

3

x

3<

x

3

> 0

limx9

x 9 3. Assuming you can choose

Hence, for you have

fx L <

x 3 <

x 9

< 5 0limx5

9 9. 12. limy4

3y 1 34 1 9

14. limt3

t2 9

t 3

limt3t

3

6 16.

limx0

1

4 x 2

1

4

limx0

4 x 2

x

limx0

4 x 2

x4 x 2

4 x 2

18.

lims0

11 s 1

s11 s 1 lim

s0

1

1 s11 s 1

1

2

lims0

11 s 1

s lim

s011 s 1s

11 s 111 s 1

20.

4

12

1

3

limx2

x 2

x2 2x 4

limx2

x2 4

x3

8

limx2

x 2x 2

x 2x2

2x 4

22. limx

4

4x

tanx

44

1

324 Chapter 1 Limits and Their Properties

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24.

0 01 0

limx0

cos x 1x limx0 sin sin x

x

limx0

cos x 1

x lim

x0coscos x sin sin x 1

x

26. limxc

fx 2gx 34 2

23

712

28.

(a)

Actual limit is

(c)

1

3

limx1

1

1 3x 3x2

limx1

1 x

x 11 3x 3x2

limx1

1 3

xx 1 lim

x1 1 3

xx 1 1 3

x

3

x

2

1 3x 3x2

13 .limx1

1 3x

x 1 0.333

fx 1 3x

x 1

(b)

3

3

3

2

x 1.1 1.01 1.001 1.0001

0.33330.33320.33220.3228fx

30.

When the velocity is approximately

limt6.39

4.96.39 6.39 62.6 msec.

limta

sa st

a t lim

ta4.9a 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 3

atx 4.

limx4

x 1 34. limx1

gx 1 1 2.

36. lims2

fs 2 38.

Removable discontinuity at

Continuous on , 1 1,x 1

limx1

3x 2 5 0

limx1

fx limx1

3x2

x 2x 1

fx 3x2 x 2x 1

,

0,

x 1

x 1

Review Exercises for Chapter 1 325

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40.

Nonremovable discontinuity at

Continuous on , 2 2,

x 2

limx2

2x 3 1

limx2

5 x 3

fx 5 x,2x 3,x 2

x > 242.

Domain:

Nonremovable discontinuity at

Continuous on ,1 0,x 0

,1, 0,

limx0

1 1x

fx x 1x 1 1

x

44.

Removable discontinuity at

Continuous on , 1 1,x 1

limx1

x 1

2x 1

1

2

fx x 1

2x 246.

Nonremovable discontinuities when

Continuous on

for all integers n.

2n 14 ,2n 1

4

x 2n 1

4

fx 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.

Chas a nonremovable discontinuity at each integer.

0

0

5

30

9.80 2.50x 1

C 9.80 2.50x 1, x > 0

54.

Vertical asymptotes at andx 2x 2

hx 4x

4 x256.

Vertical asymptote at every integer k

fx csc x

58. limx12

x

2x 1 60. lim

x1

x 1

x4 1 lim

x1

1

x2 1x 1

1

4

62. limx1

x2 2x 1

x 1 64. lim

x2

13x2 4

52.

(a) Domain:

(b)

(c) limx1

fx 0

limx0

fx 0

, 0 1,

fx x 1x

68. limx0

cos2x

x 66. lim

x0

secx

x


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Problem Solving for Chapter 1

2. (a)

(b)

(c) limx0

ax limx0

x 0

ax Area PBO

Area PAO

x22x2

x

Area PBO 1

2bh

1

21y

y

2

x2

2

AreaPAO 1

2bh

1

21x

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 atx 0.fx

fx tan 2x

x,

2,

x 0

x 0

limx0

tan 2x

x 2

fx tan 2x

x

x 0.1 0.01 0.001 0.001 0.01 0.1

2.0271 2.0003 2.0000 2.0000 2.0003 2.0271fx

4. (a)

(b) Tangent line:

(c) Let

(d)

This is the slope of the tangent line at P.

limx3

3 x

25 x2 4

6

4 4

3

4

limx3

3 x3 x

x 325 x2 4

limx3

25 x2 16

x 325 x2 4

limx3

mx limx3

25 x2 4

x 325 x2 4

25 x2 4

mx 25 x2 4

x 3

Q x,y x,25 x2

y 3

4x

25

4

y 4 3

4x 3Slope

3

4

Slope 4 0

3 0

4

3

6.

Letting simplifies the numerator.

Thus,

Setting you obtain

Thus, and b 6.a 3

b 6.b

3 33,

limx0

b

3 bx 3.

limx0

3 bx 3

x lim

x0

bx

x3 bx 3

a 3

a bx 3

xa bx 3

a bx 3

xa bx 3

xa bx 3

a bx 38.

Thus,

a 1, 2

a 2a 1 0

a2 a 2 0

a2 2 a

because limx0

tanx

x 1lim

x0fx lim

x0

ax

tanx a

limx0

fx limx0

a2 2 a2 2

Problem Solving for Chapter 1 327

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12. (a)

Let

(b)

Let

(c)

Let

Since this is smaller than the escape velocity for earth,

the mass is less.

v06.99 2.64 misec.

limv0

r10,600

6.99 v02

r10,600

v2 v02 6.99

v02.17 misec 1.47 misec.

limv0

r1920

2.17 v02

r1920

v2 v02 2.17

1920

r v2 v

02 2.17

v2 1920

r v

02 2.17

v048 43 feetsec.

limv0

r192,000

48 v02

r192,000

v v02 48

192,000

r v2 v

02 48

v2 192,000

r v

0

2 48

10.

11

1

2

3

2

1

y

x

(a)

f1 1 1

f3 13 0

f14 4 4 (b)

limx0

fx

limx0

fx

limx1

fx 0

limx1

fx 1 (c) fis continuous for allreal 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

fax limx0

f0 2.

limx0

fx 1 L,

fx

1

2

x 0

x 0.

fax L < .ax < 1

x 0 > 0a 0


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