Fisika
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...
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