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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
326 Chapter 1 Limits and Their Properties
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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
328 Chapter 1 Limits and Their Properties