Thomas Calculus 13th [Solutions]

104 Chapter 2 Limits and Continuity
6.
2 lim x lim g( x) lim x lim
x 4 x 0 x 4 x 4
constant) lim g ( x ) 2 1
0
4 2
.
x
lim g( x) 4 lim lim g( x) 4 lim g( x ) (since
x 0 x 4 x 0 x 0
lim
x 0
g( x ) is a
7. (a)
(b)
1/3 1/3
lim f ( x) lim x c f ( c ) for every real number c f is continuous on ( , ).
x c x c
3/4 3/4
lim g( x) lim x c g( c ) for every nonnegative real number c g is continuous on [0, ).
x c x c
2/3 1
c
(c)
2/3
lim h ( x ) lim x h ( c ) for every nonzero real number c h is continuous on ( , 0) and
x c x c
( , ).
1/6
lim k ( x ) lim x 1
k ( c ) for every positive real number c k is continuous on (0, )
(d)
1/6
x c x c
c
8. (a) n 1 , n 1 , where I the set of all integers.
2 2
n I
(b) ( n , ( n 1) ), where I the set of all integers.
n I
(c) ( , ) ( , )
(d) ( , 0) (0, )
9. (a)
(b)
10. (a)
(b)
2
lim x 4x 4 lim ( x 2)( x 2) 2
3 2
0 5 14 0
( 7)( 2) lim x
0
( 7)
, x
x x x x x
2; the limit does not exist because
x x x x x x
lim x 2 and lim x 2
x( x 7)
x( x 7)
x 0
x 0
lim
2
x 4x 4 lim ( x 2)( x 2) 2
3 2
2 5 14 2
( 7)( 2) lim x
2
( 7)
, x
x x x x x
2, and lim x 2 0
x x x x x x
2
x( x 7) 2(9)
0
x
2
( 1)
lim x x
x x
lim
lim x 1 lim 1 , x 0 and x 1.
5 4 3 3 2
2 2
x 0 x 2 x x x 0 x ( x 2x
1) x 0 x ( x 1)( x 1) x 0 x ( x 1)
2
Now 1 and lim 1 lim x x .
2
xlim
0 x ( x
2 5 4 3
1) x 0 x ( x 1) x 0 x 2x x
2
( 1)
lim x x
x x
lim lim 1 , x 0 and x 1. The limit does not exist because
5 4 3 3 2 2
x 1 x 2 x x x 1 x ( x 2x 1) x 1 x ( x 1)
lim 1 and lim 1 .
2
x 1 x ( x
2
1)
x 1 x ( x 1)
11.
12.
13.
14.
15.
16.
1 x
1 x
lim lim lim 1 1
x 1
1 x
x 1 (1 x )(1 x) x 1 1 x 2
2 2
2 2
( )
lim x a
x a
lim lim 1 1
x a x a x a ( x a )( x a ) x a x a 2a
4 4 2 2 2 2 2 2 2
2 2 2 2 2
lim ( x h) x lim ( x 2 hx h ) x
lim (2 x h ) 2 x
h 0
h
h 0
h
h 0
2 2 2 2 2
lim ( x h) x lim ( x 2 hx h ) x
lim (2 x h ) h
x 0
h
x 0
h
x 0
1 1
x
2 2 2 (2 x) lim lim lim 1 1
x 0
x
x 0
2 x(2 x) x 0
4 2x
4
3 3 2
lim (2 x) 8 lim ( x 6x 12x
8) 8 lim ( 2
x 6 x 12) 12
x 0
x
x 0
x
x 0
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