Thomas Calculus 13th [Solutions]

Chapter 2 Additional and Advanced Exercises 113
1 1 a 1 1 a 1 1 a
1 (1 a)
(b) At x 0: lim r ( a) lim lim
lim
lim a
a a
a 0 a 0 a 0
1 1 a a 0 a( 1 1 a) a 0 a( 1 1 a)
lim 1 (because the denominator is always negative); lim r ( a)
a 0 1 1 a
a 0
lim 1 (because the denominator is always positive).
a 0 1 1 a
Therefore, lim r ( a) does not exist.
a 0
1 1
At 1: lim ( ) lim a
x r a
1
a
lim 1
a 1 a 1 a 1 1 1 a
(c)
(d)
22. f ( x) x 2 cos x f (0) 0 2 cos 0 2 0 and f ( ) 2 cos( ) 2 0. Since f ( x ) is
continuous on [ , 0], by the Intermediate Value Theorem, f ( x ) must take on every value between [ 2, 2].
Thus there is some number c in [ , 0] such that f ( c ) 0; i.e., c is a solution to x 2 cos x 0.
23. (a) The function f is bounded on D if f ( x)
M and f ( x)
N for all x in D. This means M f ( x)
N for
all x in D. Choose B to be max {| M |, | N |}. Then | f ( x)| B . On the other hand, if | f ( x)| B , then
B f ( x) B f ( x)
B and f ( x) B f ( x ) is bounded on D with N B an upper bound and
M B a lower bound.
(b) Assume f ( x)
N for all x and that L N . Let L N
2 . Since lim f ( x)
L there is a 0 such that
x x0
0 | x x0
| | f ( x) L|
L f ( x) L L L N f ( x) L L N L N f ( x)
2 2 2
3L N . But L N L N N N f ( x ) contrary to the boundedness assumption f ( x) N . This
2
2
contradiction proves L N.
(c) Assume M f ( x ) for all x and that L M . Let M L .
2
As in part (b), 0 | x x 0 | L M L
3
2
f ( x) L M L L M f ( x) M L M , a contradiction.
2 2 2
Copyright
2014 Pearson Education, Inc.