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

Special Kinds of Theorems 101
that lim a„ ^ L, one needs to prove that there exists an £ > 0 such that for all
numbers N there is an n>N such that |a„ — L\ > 8. In this statement, Lis
a real number.
To prove that lim a„ does not exist, one needs to prove that the
statement above is true for all real numbers; that is, lim an^L
numbers L
EXAMPLE 6.
does not exist.
n->oo
for all real
Consider the sequence [an = (-l)"}^!- Prove that its limit
Proof: To show that the conclusion is true we need to prove that there
exists an 6:>0 such that for all numbers N there is an n>N such that
\(in— L\ > 6, where Lis any real number.
(As the values of the terms of the sequence oscillate between -1 and 1, it
might make sense to use a value of ^ smaller than 1.) Consider s = 9/10. We
have to consider two possible cases for L: L < 0 and L>0.
Case 1. L < 0. (We want to prove that the terms of the sequence are
somewhat "far" from L. As Lis nonpositive, and some of the terms of the
sequence are positive, we could try to use them to reach the goal.). Let N be
any number and let t be an even number, where t>N. Then at = (—1)^ = L
So, \at — L\ = [1 — L|. Because —L > 0, 1 — L > 1. Therefore,
|af-L| = |1-L| =: 1-L> 1 >£.
Case 2. L > 0. (We want to prove that the terms of the sequence are
somewhat "far" from L Because Lis positive and some of the terms of the
sequence are negative, we could try to use them to reach the goal.). Let N be
any number and let s be an odd number, where s>N. Then as = (—1)^ = — 1.
So,
\as - L| =3 1-1 - L[ = |-(1 + L)| = 1 + L > 1 > f.
Thus, the sequence does not have a limit.
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EXERCISES
1. Prove that lim Sx^ + 2 = 5.
2. Prove that Hm -iz—- = -
x->2 X^ + 1 5