1549901369-Elements_of_Real_Analysis__Denlinger_

182 Chapter 4 • Limits of Functions

Solution: (a) We want a real number o > 0 such that 0 < Ix - 21 < o::::}

lx^2 - 41 < .01. That is, we want to have

-.01 < x^2 - 4 <. 01

3.99 < x^2 < 4.01

J3]9 < x < J4.01

1.99749 ... < x < 2.0024 ... '

which will be true if 1.998 < x < 2.002; that is,

2 - .002 < x < 2 + .002

Ix - 21 < .002.

Thus, we take o = .002.

(b) We want a real number o > 0 such that 0 < Ix - 21 < o::::} lx^2 - 41 =
Ix+ 2llx - 21 < c:. The reasoning used here will probably be new to you; follow
it closely. First, suppose we have o ::::; l. Then

Ix - 21 < o ::::} Ix - 21 < 1

::::} -1 < x - 2 < 1

::::}l<x<3

::::}3<x+2<5

::::} Ix+ 21<5.

Thus, when Ix - 21<1, we will have Ix+ 21 < 5. Remember that we want

lx+2llx-21 <c:.

c:

Thus, in addition to Ix - 21 < 1, we want Ix - 21 < S in order to guarantee

that Ix+ 2llx - 21 < 5 · ~ = c:. Therefore, we take o = min{l, H· D

In the next example we show how to use the result of Example 4.1.5 (b)

to prove that lim x^2 = 4. As we did in Chapter 2, we regard the work done in

X->2

Example 4.1.5 (b) as "scratchwork,'' and work backwards to produce our proof.