1549901369-Elements_of_Real_Analysis__Denlinger_

204 Chapter 4 11 Limits of Functions

lx-21
Example 4.3.2 Prove that lim --
2

= -1.

x_,2- X -

y

-l+e

Given e

-1-e ':_1

lx-2 1

y=--

x-2

2- s

\~o>O

2

Figure 4.7

x

Proof. Let c > 0. Let o = any positive number. Then, 2 - o < x < 2 :;.

I

Ix -

2

1 - (-1)1 = 1-(x -

2

) - (-1)1 = 1(-1) - (-1)1 = 0 < €. Therefore,

x-2 x-2

lim lx-21=-l. D

x_,2- X - 2

Definition 4.3.3 (Limit from the Right) If x 0 is a cluster point of D(j) n

(x 0 , oo), then the limit off from the right is L , written f(xci) = lim J(x) = L ,

X-tXt

if

I Ve> 0, :3 o > 0 3 Vx E D(j), xo < x < xo + o:;. lf(x) - LI < c. j

Notes on Definition 4.3.3:

(1) We shall never say that lim f(x) exists unless x 0 is a cluster point

X-+Xo+

of D(j) n (x 0 , oo).

(2) Even if xo E D(f), the value of f(xo) is irrelevant to the consideration

of whether lim f(x) = L. The condition "xo < x < Xo + o" in Definition 4.3.3

X---tXt

guarantees that, when we consider whether lim f(x) = L , we are never letting

x-+xci

X = Xo.

(3) If D(f) contains some interval of the form (x 0 , xo +1), for some r > 0,

then Definition 4.3.3 simplifies to:

lim f ( x) = L if V c > 0, :3 o > 0 3 xo < x < xo + o :;. If ( x) - LI < c.

X-+xci