Thomas Calculus 13th [Solutions]

96 Chapter 2 Limits and Continuity
57. (a)
(b)
(c)
(d)
(e)
58. (a)
(b)
(c)
(d)
(e)
2
( 2)( 1) negative negative
lim x 3x
2 lim x x
3 2 2
2 ( 2)
positive negative
x 0 x x x 0 x x
2
( 2)( 1)
lim x 3x 2 x x
lim lim x 1 1 , x
3 2 2 2
2 ( 2)
4
x 2 x x x 2 x x x 2 x
2
2
( 2)( 1)
lim x 3x 2 x x
lim lim x 1 1 , x
3 2 2 2
2 ( 2)
4
x 2 x x x 2 x x x 2 x
2
2
( 2)( 1)
lim x 3x 2 x x
lim lim x 1 1 , x
3 2 2 2
x 2 x 2 x x 2 x ( x 2) x 2 x 4
2
( 2)( 1) negative negative
0 x 2 x x 0 x ( x 2)
positive negative
2
lim x 3x
2 lim x x
3 2 2
x
2
( 2)( 1) ( 1)
lim x 3x
2 x x x
lim lim
1 1
3
4
x( x 2)( x 2) x( x 2) 2(4) 8
x 2 x x x 2 x 2
lim
2
x 3x
2 lim ( x 2)( x 1) ( 1) negative
3
4
( 2)( 2) lim x
x x x x( x 2) negative positive
x 2 x x x 2 x 2
lim
2
x 3x
2 lim ( x 2)( x 1) ( 1) negative
3
4
( 2)( 2) lim x
x x x x( x 2) negative positive
x 0 x x x 0 x 0
2
( 2)( 1) ( 1)
lim x 3x
2 x x x
lim lim 0 0
3
4
x( x 2)( x 2) x( x 2) (1)(3)
x 1 x x x 1 x 1
1
negative
lim x
x( x 2) positive positive
x 0
negative
and lim x 1
x( x 2) negative positive
x 0
so the function has no limit as x 0.
59. (a) 2 3
1/3
tlim
0 t
60. (a) lim 1 7
3/5
t 0 t
61. (a) 1 2
2/3 2/3
xlim
0 x ( x 1)
(c) 1 2
2/3 2/3
xlim
1 x ( x 1)
62. (a) 1 1
1/3 4/3
xlim
0 x ( x 1)
(c) 1 1
1/3 4/3
xlim
1 x ( x 1)
(b) 2 3
1/3
tlim
0 t
lim 1 7
t 0 t
(b)
3/5
(b) 1 2
2/3 2/3
xlim
0 x ( x 1)
(d) 1 2
2/3 2/3
xlim
1 x ( x 1)
(b) 1 1
1/3 4/3
xlim
0 x ( x 1)
(d) 1 1
1/3 4/3
xlim
1 x ( x 1)
63. y 1
x 1
64. y 1
x 1
Copyright
2014 Pearson Education, Inc.