Thomas Calculus 13th [Solutions]

Chapter 2 Practice Exercises 105
17.
lim 1/3
1/3 2/3 1/3
x 1 lim x x x x
2/3 1/3
x
( 1) ( 1)( 1)
1 x 1 x 1 ( x 1) ( x 1)( x x 1)
( x 1)( x 1) x
lim
lim
1 1 1 2
2/3 1/3 2/3 1/3
x 1 ( x 1)( x x 1) x 1 x x 1 1 1 1 3
18.
1/3 1/3
1/3 1/3 2/3 1/3
lim 2/3
x 16 lim ( x 4)( x 4) ( 4)( 4) ( 4 16)( 8)
lim x x x x x
2/3 1/3
x 64 x 8 x 64 x 8 x 64 x 8 ( x 8)( x 4x
16)
( 1/3 1/3
x 64) ( x 4) ( x 8) ( x 4) ( x
lim
lim
8) (4 4) (8 8) 8
2/3 1/3 2/3 1/3
x 64 ( x 64) ( x 4x 16) x 64 x 4x
16 16 16 16 3
19.
tan 2x
sin 2 cos
lim lim x x
x 0
tan x
x 0
cos 2x sin x
lim sin 2x cos x x 2x
2 2
0
2x cos 2x sin x x
1 1 1
x
20. lim csc x lim 1
sin x
x
x
21.
22.
lim sin x
x
2
sin x sin
2
sin sin
2
1
2
lim cos ( x tan x) 2
cos ( tan )
2
cos ( )
2
( 1) 1
x
lim 8x
8 8
3sin lim 3 1 3(1) 1
4
23.
sin x
x 0
x x
x 0 x
24. lim cos 2x 1 lim cos 2x 1 cos 2 x 1
x 0
sin x
x 0
sin x cos 2x
1
lim
2 2
4sin x cos x 4(0)(1)
0
cos 2x
1 1 1
0
x
2 2
lim cos 2x
1 lim sin 2x
x 0
sin x(cos 2x 1)
x 0
sin x(cos 2x
1)
25. Let x = t 3
lim ln( t 3) lim ln x
t 3 x 0
26.
27.
2
lim t ln 2 t ln1 0
t 1
1 cos 1 e
1 cos( / )
e
1
e e
1 cos( / )
e e lim
cos( / )
e 0 by the
0
Sandwich Theorem
1/ z
2e
28. lim lim 2 2 2
1/ 1/ z
1 1 0
z 0
z
e 1 z e
29.
1/3
1/3
lim [4 g( x)] 2 lim 4 g( x) 2 lim 4 g( x ) 8, since 2 3 8. Then lim g( x) 2.
x 0 x 0 x 0
x 0
30. lim 1 2 lim ( x g( x))
1
1 1
x 5
x g( x) x 5
2
5 lim g ( x )
5
2 lim g ( x )
5
2
5
x
x
31.
32.
2
2
lim 3x
1
1
( )
lim g ( x ) 0 since lim (3 x
g x
1) 4
x x 1 x 1
2
2
lim 5 x 0 lim g ( x ) since lim (5 x ) 1
x 2 g( x)
x 2 x 2
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