1549901369-Elements_of_Real_Analysis__Denlinger_

4.4 *Infinity in Limits 219

Proof. (a) Part 1 (=>):Suppose lim j(x) = L. Let c > 0. Then 38 > 0 3

X-+O+

1

0 < x < 8 => If ( x) - LI < c. Let M = J. Then

x > M 0 < ~ < ~ = 8 It (~)-LI<€.

Therefore, lim f (~) = L.
x-++oo X

Part 2 ({::::):Suppose lim t(~) = L. Then 3M > 0 3 x > M =>

x-++oo X

If ( ~) -LI < c. Let 8 = ~. Then 8 > O and

1 1

0 < x < 8 => x < M =>; > M => lf(x) - LI< c.

Therefore, lim f(x) = L.
x-+O+

To prove (b ), modify the proof of (a) in the obvious ways. •

Note: From Theorem 4.4.19 it follows that lim ~ = 0, because lim x = 0.
x-++oo X x-+O+
1
Similarly, lim - = 0, because lim x = 0.
X-+-oo X x-+O-

Many other results like these follow from another, similar theorem. First,
however, we make a useful definition.

Definition 4.4.20 (a) A neighborhood of +oo is any open interval of the
form (a, +oo).
(b) A neighborhood of -oo is any open interval of the form (-oo, a).

Theorem 4.4.21 (a) Suppose f(x) > 0 for all x in some neighborhood of +oo.
Then

x-++oo lim f(x) = +oo {:} x-++oo lim f(l X ) = 0.

(b) Suppose f(x) < 0 for all x in some neighborhood of +oo. Then

x-++oo lim f(x) = -oo {:} x-++oo lim f(l) X = 0.

(c) Suppose f(x) > 0 for all x in some neighborhood of -oo. Then

1

X-+-00 lim f(x) = +oo {:} X-+-00 lim -f( X ) = 0.