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