210 Chapter 4 a Limits of Functions
1
(b) For arbitrary M > 0, find 8 > 0 3 0 < Ix - 21 < 8 => (x _
2
) 2 > M.Solution. (a) We want to find a 8 > 0 such that 0 < Ix - 21 < 8 =>
(x ~
2
) 2 > 1,000. Equivalently, (x-2)^2 < 10100 = .001. Thus, we want lx-21 <V,06I = 0.031632 · · ·. So, we take 8 = 0.03163, or even 8 = 0.03.
1
(b) We want to find a 8 > 0 such that 0 < Ix - 21 < 8 => (x _
2
) 2 > M.We want (x - 2)^2 < ~. Equiva lently, Ix - 21 < .j"k. We take 8 = .j"k. D
In general, using Definition 4.4.1 to prove that lim f(x) = +oo is more
X--+Xo
difficult than Example 4.4.2 would lead you to believe. For an example of the
difficulty, try to prove lim /x -; 2 = +oo using only Definition 4.4.1. The
x->2 X - 2
following theorem provides a useful technique that will often be helpful.
Theorem 4.4.3 Suppose f(x) > 0 for all x in some deleted neighborhood of
1
xo. Then lim f(x) = +oo ¢::> lim f( ) = 0.
X--+Xo X--+Xo x
Proof. Suppose f(x) > 0 for all x in some deleted neighborhood of x 0.
(a) Part 1 (=> ): Suppose lim f(x) = +oo. Let E: > 0. Then ~ > 0. Since
x--+xo C
lim f(x) = +oo, 38 > 0 3 \:/x E D(f),
X--+Xo
1
0 < Ix - xol < 8 => f(x) > -
1 E:
=> f(x) < E:.Therefore, lim f(l ) = 0.
X--+Xo X
(b) Part 2 (~):Suppose lim f(l) = 0. Let M > 0. Then J_ > 0. Since
x->xo X M. 1
X->Xo hm f( X ) = 0,^3 8 > 0 3 \:/x E D(f),
1 1
0 < Ix - xol < 8 => 0 < f(x) < M
=> f(x) > M.Therefore, lim f(x) = +oo. •
X--+Xo