James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A42
appendix F Proofs of Theorems
3 The Squeeze Theorem If f sxd < tsxd < hsxd for all x in an open interval that
contains a (except possibly at a) and
lim f sxd − lim hsxd − L
x l a x l a
then
lim tsxd − L
x l a
ProoF Let « . 0 be given. Since lim x l a f sxd − L, there is a number 1 . 0 such that
that is,
if 0 , | x 2 a | , 1 then | f sxd 2 L | , «
if 0 , | x 2 a | , 1 then L 2 « , f sxd , L 1 «
Since lim x l a hsxd − L, there is a number 2 . 0 such that
that is,
if 0 , | x 2 a | , 2 then | hsxd 2 L | , «
if 0 , | x 2 a | , 2 then L 2 « , hsxd , L 1 «
Let − minh 1 , 2 j. If 0 , | x 2 a | , , then 0 , | x 2 a | , 1 and
0 , | x 2 a | , 2, so
L 2 « , f sxd < tsxd < hsxd , L 1 «
In particular, L 2 « , tsxd , L 1 «
and so | tsxd 2 L | , «. Therefore lim x l a tsxd − L.
■
Section 2.5
Theorem If f is a one-to-one continuous function defined on an interval sa, bd,
then its inverse function f 21 is also continuous.
prooF First we show that if f is both one-to-one and continuous on sa, bd, then it must
be either increasing or decreasing on sa, bd. If it were neither increasing nor decreasing,
then there would exist numbers x 1 , x 2 , and x 3 in sa, bd with x 1 , x 2 , x 3 such that
f sx 2 d does not lie between f sx 1 d and f sx 3 d. There are two possibilities: either (1) f sx 3 d
lies between f sx 1 d and f sx 2 d or (2) f sx 1 d lies between f sx 2 d and f sx 3 d. (Draw a picture.)
In case (1) we apply the Intermediate Value Theorem to the continuous function
f to get a number c between x 1 and x 2 such that f scd − f sx 3 d. In case (2) the Intermediate
Value Theorem gives a number c between x 2 and x 3 such that f scd − f sx 1 d. In either
case we have contradicted the fact that f is one-to-one.
Let us assume, for the sake of definiteness, that f is increasing on sa, bd. We take
any number y 0 in the domain of f 21 and we let f 21 sy 0 d − x 0 ; that is, x 0 is the number
in sa, bd such that f sx 0 d − y 0 . To show that f 21 is continuous at y 0 we take any « . 0
such that the interval sx 0 2 «, x 0 1 «d is contained in the interval sa, bd. Since f is
increasing, it maps the numbers in the interval sx 0 2 «, x 0 1 «d onto the numbers in the
interval s f sx 0 2 «d, f sx 0 1 «dd and f 21 reverses the correspondence. If we let denote
the smaller of the numbers 1 − y 0 2 f sx 0 2 «d and 2 − f sx 0 1 «d 2 y 0 , then the
interval sy 0 2 , y 0 1 d is contained in the interval s f sx 0 2 «d, f sx 0 1 «dd and so is
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