1549901369-Elements_of_Real_Analysis__Denlinger_

4.4 *Infinity in Limits 211

Example 4.4.4 below shows the power of using Theorem 4.4.3 in proving

that lim f(x) = +oo.

X-+XQ

3x-5

Example 4.4.4 Prove that X->2 lim ( X - 2 ) 2 = +oo.

Proof. By the algebra of limits, established in Section 4.2, we have

. (x - 2)^2 0
hm
3

= - = 0. Moreover, as x ---+ 2, 3x - 5 ---+ 1, so for x sufficiently

X->2 X - 5 1

3x - 5 3x - 5

close to 2, (x _ 2 ) 2 > 0. Thus, by Theorem 4.4.3, ~~ (x _

2

) 2 = +oo. D

INFINITY AS A ONE-SIDED LIMIT

Definit ion 4.4.l can be altered to define lim f(x) = + oo, lim_ f(x) =

X-+XQ X-+x 0

• oo, lim f(x) = +oo, and lim f(x ) = -oo. (Exercise 3)
  x -+xci x---+xci

x-2 x-2
Example 4.4.5 Investigate li m --and lim --
x_,1-x - l x->I+x-1

Solution: In Exercise 4, we modify Theorem 4.4.3 to cover one-sided limits.
We show here how to apply these simple modifications.
x -1 0
(a) First, observe that lim --= - = 0. Moreover, as x---+ 1-, x < 1
X->l-X-2 -1
x-1
and x < 2, so x -1 < 0 and x - 2 < O; t hus as x---+ 1-, --> 0. Thus, using
x-2
x-2
Exercise 4, lim - - = +oo.
X->l -X - 1
x -1 0
(b) Next, observe that lim --= - = 0. Moreover, as x ---+ 1 + , x > 1
x->l+ X - 2 -1
x- 1
and x < 2, so x - 1 > 0 and x - 2 < O; thus as x ---+ 1 + , --
2

< 0. Thus, using
x -
x-2
Exercise 4, lim --= -oo. D
X->l + X - 1

Theorem 4.4.6 If x 0 is a cluster point of D(f) n( - oo, xo), and a cluster point
of D(f) n(xo, oo), then

(a) lim f(x) = +oo ¢:? both lim f(x) = +oo and lim f(x) = +oo;

x -+xo x---+xQ x-+xri

(b) lim f(x) = -oo ¢:? both lim f(x) = -oo and lim f(x) = -oo.

x---+xo x-+x() x---+xci

Proof. Exercise 7. •