epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Special Kinds of Theorems 95
The existence of 8 will be proved explicitly, by providing a formula for it
that depends on s and possibly on 1.
Note that numbers —1 and 1 are not in the domain of the function
f(x) = (x — i)/(x^ — 1). So, in particular, /(x) is not defined at c = 1 (see
Fact 3 above). While this might make the algebraic steps more delicate to
handle, the logic steps will be similar to those in Example 1.
Let 6:>0 be given. How close to 1 will x have to be for the inequality
|((x - l)/(x^ - 1)) - (1/2) I <sto hold true? Let us start by simplifying the
expression in the absolute value:
x-l
c2-l
1
x+1
1-x
2(x+l)
2-(x+l)
2(x + l)
II -x\
2|x+l|
By the properties of the absolute value function |1 — x| = |x — 1|. Thus,
1
x2-l
1
11
2ix+ir
Once more, we need to prove that there exists a positive number 8
such that, if \x-l\<8, then the quantity \{(x-l)/(x^ - I)-{l/2))\ =
\x — l|(l/(2|x + 1|)) will be smaller than the given 6>0.
What is the largest value that the fraction can have for x
2|x + l|
sufficiently close to 1?
Because the fraction is undefined for x = —1, we should avoid this value.
Thus, let us choose an interval centered at 1 that does not include it, such
1 9
as — 4 T < X < -. This is an interval with center at 1 and radius r = 5/4
{i.e., \x 1| < 5/4). (The choice of the radius is arbitrary; we can choose any
positive number smaller than 2 to exclude x =
the following estimates for the fraction
2|x+l|
1 9
4 4
3 , 13
- < x-f 1 < —
4 4
3 , 1, 13
4<|x+l|<-
-<2|x+l|
13
2_ 1
13^2|x+l|
"3
L) For r = 5/4, we obtain