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

96 The Nuts and Bolts of Proof, Third Edition
Therefore, whenever -(l/4)<x<(9/4),
Then, the quantity x-1
x2-l
that is, if \x
|x-l 1
x2-l 2 = |x-l| 1
2|x+l| <
l\<ls.
Let us choose 8 = minimum
X •
will be smaller than s for sure if |x — 11 - < e;
3 m •"
X- 1\<S < 5/4 and |x - 1| <3 < -s, we will have
this way,
x-1 1,
because
Therefore, we have proved that for every s>0 there exists a
f5 3 1
8>0 (namely, 8 = minimum { T , x ^ [), such that if 1|<5,
x-1 11 „. 14 2 J
then
<6. This proves that Hm -^^^ = k
1 2
It is possible to check whether the result obtained is indeed correct, and it
should be emphasized that some mathematicians consider this checking
process to be an important part of the proof.
In Example 2, let x satisfy the inequality \x—\\<8 with
8 = minimum{(5/4),(3/2)^} (so, in particular, 8 < 5/4 and 8 < (3/2)e). Will
lx-1 11 _
it really follow that
1 2 \<sl
x-1
1
As seen before.
Thus, we need to consider
I 2|x+l|
the two factors |x — 1| and , -. We already have an estimate for the
2|x+l| 5
first, as |x - 1| <5. What about the second? Because |x - 1| <5< -, it follows
that:
5 . 5
4 4
1 9
4 4
3 , 13
-< x-hl < —
4 4
3 , 1, 13
-<|x+l|<-
3 13
^<2|x + l|<y
13
1 2
2|x+l| ^3*