196 Chapter 4 • Limits of Functions

Proof. Exercise 14. •

The following example shows how Theorem 4.2.18 is used in practice.

.. 2x^2 - 18
Example 4.2.19 Fmd hm
x-+3 X - 3
. 2x^2 - 18 2(x - 3)(x + 3)
Solution: For x -j. 3,
3

3
= 2x + 6. Thus, by
x- x-
. 2x^2 - 18.
Theorem 4.2.18, hm = hm (2x + 6) = 12. D
x-+3 X - 3 x -+3

INEQUALITIES AND LIMITS

Theorem 4.2.20 (The "Squeeze" Principle for Functions)

(a) The First Squeeze Principle: Suppose f(x) ~ g(x) ~ h(x) for all

x in some deleted nbd. of xo, and lim f(x) = lim h(x) = L. Then

x--+xo x--+xo

lim g(x) = L.

X--+Xo

(b) The Second Squeeze Principle: Suppose lim g(x) = 0. If lf(x)-LI ~

X-+Xo

lg(x)I, for all x in some deleted nbd. of xo, then lim f(x) = L.

X--+Xo

Proof. (a) Suppose f(x) ~ g(x) ~ h(x) for all x in some deleted nbd.
of xo, and X--+XQ lim f(x) = X--J-XQ lim h(x) = L. Then, ::J 81 > 0 3 'Vx E N8 1 (xo),

f(x) ~ g(x) ~ h(x). Let E. > 0.

Since lim f(x) = L, ::l 82 > 0 3 'Vx E N8 (xo), lf(x) - LI < E..

X--+Xo 2

Since X--+Xlim Q h(x) = L, :383 > 0 3 'Vx E N8 3 (xo), lh(x) - LI< E..

Let 8 = min{8 1 ,82,83}. Then

0 < Ix -xo I < 8 =? 0 < Ix -xo I < 81 , 82 and 83

=? f(x) ~ g(x) ~ h(x), lf(x) - LI < E. and lh(x) - LI< E.

=? f(x) ~ g(x) ~ h(x), L - E. < f(x) and h(x) < L + E.

=? L - E. < f(x) ~ g(x) ~ h(x) < L + E.

=? L - E. < g(x) < L + E. =? lg(x) - LI < E..

Therefore, lim g(x) = L.

X--+Xo