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C H A P T E R 1 Limits and Their Properties
Section 1.1 Section 1.2 Section 1.3 Section 1.4 Section 1.5 A Preview of Calculus . . . . . . . . . . . . . . . . . . . 305 Finding Limits Graphically and Numerically . . . . . . . 305 Evaluating Limits Analytically . . . . . . . . . . . . . . . 309 Continuity and One-Sided Limits Infinite Limits . . . . . . . . . . . . . 315

. . . . . . . . . . . .. . . . . . . . . . . 320

Review Exercises

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324

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

C H A P T E R 1 Limits and Their Properties
Section 1.1 A Preview of Calculus
Solutions to Even-Numbered Exercises
2. Calculus: velocity is not constant Distance 20 ft sec 15 seconds
2

4.Precalculus: rate of change 300 feet

slope

0.08

6. Precalculus: Area 2 5 2 5 3 5 1.5

2

8. Precalculus: Volume

3 26

54

10. (a) Area Area

5 1 5 2

5 4 5 2

10.417 5 2.5 5 3 5 3.5 5 4 5 4.5 9.145

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

Section 1.2
2. x f x lim 1.9 0.2564 x x2 2 4

Finding Limits Graphically and Numerically
1.99 0.2506 0.251.999 0.2501 2.001 0.2499 2.01 0.2494 2.1 0.2439

x→2

Actual limit is 1 . 4

4.

x f x

3.1 0.2485 1 x x 2 3

3.01 0.2498

3.001 0.2500

2.999 0.2500
1 4.

2.99 0.2502

2.9 0.2516

x→ 3

lim

0.25

Actual limit is

6.

x f x

3.9 0.0408 x x x 1

3.99 0.0401 4 5 4

3.999 0.0400

4.001 0.0400

4.01 0.0399
1

4.1 0.0392

x→4

lim

0.04

Actuallimit is 25 .

8.

x
f x lim

0.1 0.0500
cos x x 1

0.01 0.0050
0.0000

0.001 0.0005

0.001 0.0005

0.01 0.0050

0.1 0.0500

x→0

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

305

306

Chapter 1

Limits and Their Properties lim x2

10. lim x2
x→1

2

3

12. lim f x
x→1

x→1

2

3

14. lim

1 does not exist since the x 3 function increasesand decreases without bound as x approaches 3.
x→3

16. lim sec x
x→0

1

18. lim sin x
x→1

0

20. C t (a)
1

0.35

0.12

t

1

0 0

5

(b)

t Ct lim C t t Ct

3 0.59

3.3 0.71

3.4 0.71

3.5 0.71

3.6 0.71

3.7 0.71

4 0.71

t→3.5

0.71 3 0.47 2.5 0.59 2.9 0.59 3 0.59 3.1 0.71 3.5 0.71 4 0.71 3.

(c)

t→3.5

lim C t does not exist. Thevalues of C jump from 0.59 to 0.71 at t

22. You need to find such that 0 < x 2 < implies f x 3 x2 1 3 x2 4 < 0.2. That is, 0.2 4 0.2 3.8 3.8 3.8 2 So take Then 0 < x 4.2 3.8
< x2 4 < x2 < x2 < x < x 2 < 0.2 < 4 0.2 < 4.2 < 4.2 < 4.2 2

24. lim 4
x→4

x 2 x 2 2

2 2 < 0.01 x < 0.01 2 4 < 0.01

4

4.2 2 <

2

0.0494. implies 2 < 2 < 4.2 4.2 2 2.

1 x 2 0 < x

4 < 0.02 4 <
< 0.012 < x 2 < x

Hence, if 0 < x 1 x 2 2 4 x 2 f x 4

0.02, you have

Using the first series of equivalent inequalities, you obtain f x 3 x2 4 < 0.2.

x < 0.01 2 2 < 0.01 L < 0.01

Section 1.2 26. lim x2
x→5 2

Finding Limits Graphically and Numerically

307

4 4 x2

29 29 < 0.01 25 < 0.01 5 < 0.01 0.01 x 5 0.01 11 0.0009.

28. lim 2x
x→ 3

5 > 0: 5 2x 2x x

1

xGiven 2x

1

<

x

5 x x

6 < 3 < 3 < 2. 3 < 3 < 6 < 1
<

5 <

If we assume 4 < x < 6, then Hence, if 0 < x x x 5 x x2 x2 4 f x 5 <

2

0.01 , you have 11

Hence, let Hence, if 0 < x x 2x 2x 5 f x

0.01 1 5 < < 0.01 11 x 5 5 < 0.01 25 < 0.01 29 < 0.01 L < 0.01

2 2

, you have

L <

30. lim

x→1

2 3x

9 > 0: 9
2 3x 2 3

2 3

1

9

29 3

32. lim

x→21 > 0:

1 1 1 < 0 <

Given
2 3x

Given
29 3 2 3

< < Hence, any

> 0 will work.
> 0, you have

x
3 2

1 <

Hence, for any 1 f x
3 2

x Hence, let

1 <

1

<

3 2 . 1 < 1 <
2 3 29 3 3 2

L <

Hence, if 0 < x x
2 3x 2 3x

, you have

< <

9 f x

L <

34. lim

x→4

x > 0:

4

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

36. lim x
x→3

3 > 0: 3 x...