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

148 The Nuts and Bolts of Proof, Third Edition
condition x € ^ is redundant when selecting the elements in the union.
Indeed, every element of A will be in the union because it is an element
of B. Thus, A\JB = {x\x eAoxxeB} = {x\x £ B] = B.
LIMITS
Let £ > 0 be given. We need to prove the existence of a number 5 > 0 such
that if \x - 1| <5, then \{3x^ + 2) - 5| <s. Because \{3x^ + 2) - 5| =
|3x^ - 3| = 3\x^ ~ l| = ^1^ + l|l^ ~ 1|' we need to estimate how large
the value of the factor |x + 1| can be. Because x has to be in an
interval centered at 1, we can choose 0<x<2 {i.e., an interval of
radius 1). (This is a completely arbitrary choice. Other choices will
work as well; they will just yield different results for <5>0. For
example, check what would happen when one uses -1 < x < 3.)
Therefore, 1 < x + 1 < 3; that is, 1 < |x + 11 < 3. Thus,
|(3x^ + 2) - 5| = 3|x + l||x - 1| < 3 X 3|x - 1| = 9|x - 1|.
Because we want 9|x-l|<e, we need |x-l|<e/9. In conclusion,
choose 8 = minimum!l,£/9}. Because £>0, we have 8>0. Remember
that it is possible to check that if |x — 1| <5, where 5>0 is the number
we determined, then |(3x^ + 2) — 5| <e.
2. Let £ > 0 be given. We need to prove the existence of a number 5 > 0
such that if |x - 2| <5, then (| l/(x^ + 1) - (1/5)| <6).
Algebraic steps yield:
1
X2 + 1
5 - (x^ + 1)
5(x2 + 1)
4-x^
5(x2 + 1)
|4-x^|
|5(x2 + l)|'
The quantity 5(x^ + 1) is always positive (because x^ + 1 is always
positive). So, 5(x^ + 1) = |5(x^ + 1)|- Moreover, we need to keep in
mind that we want to estimate the value of |x —2|. Thus, we can
consider the following equality |4 — x^l = |x^-4| = |x + 2||x--2|.
Therefore, 1 5(x2+i) 1^ ~ 2| ^^^ we need to estimate the
X2 + 1
largest value that the fraction ^j^iipk can have for x in an interval
centered at 2. Because x has to be m an interval centered at 2, we
can choose 0.5<x<3.5 (i.e., an interval of radius L5). (This is a
completely arbitrary choice. Other choices will work as well; they
will just yield different results for 8>0. For example, check what
would happen when one uses l<x<3 or 0<x<4.) Therefore,
2.5 < X + 2< 5.5; that is, 2.5 < |x + 2| < 5.5. Moreover 0.25 <x^ < 12.25.