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

Special Kinds of Theorems 97
Thus,
\x-i 11
im-
x2-l 2 2|x + l| 3
It is considerably more difficult to prove that either lim/(x) ^ L ov that
lim/(x) does not exist. This is partly due to the fact that the definition of
limit involves several quantifiers.
For example, one can state that lim/(x) / L if there exists one e > 0 for
which there is no 5 > 0 such that if |x - c| < 5, then |/(x) - L| < 6: for all x. In
this case. Lis assumed to be a real number.
This is equivalent to stating that lim/(x) / L if there exists one ^ > 0 such
x->c
that for every 8>0 there is at least one xs with \x8 — c\<8 but such that
|/(X,)-L|>£.
To prove that lim/(x) does not exist, one would have to prove that
x^-c
lim/(x) ^ L for all real numbers L
x~^c
X
EXAMPLE 3. Let f(x) = —-. Then Um /(x) does not exist.
|x| x-^O
Proof: The function/(x) is not defined at 0, but this does not necessarily
imply the nonexistence of the Umit.
(See Example 2. It might be useful to graph the function around 0 to study
its behavior. The graph will show a "vertical jump" of 2 units at 0. Moreover,
the values of/(x) are positive for positive values of x and negative for
negative values of x.)
Analytically, we need to prove that lim /(x) ^ L for all real numbers L.
x^-O
Therefore, we need to prove that it is possible to find an e>0 such that for
every 8>0 there is at least one x^ with \x8 — 0\<8 but such that
\f(X8)-L\>S.
Consider e = 1/2 (this choice is somewhat arbitrary; see the comments
that follow this proof) and let 8 be any positive number. As Lis a real
number, there are two possible cases: L < 0 and L>0.
Case 1. L < 0. (We want to prove that the values of/(x) are "not very
close" to L. Because Lis nonpositive, we can try using positive values
of/(x) that correspond to positive values of x.)
8/4 8/4
Let Xs = 8/4, so \xs -0\<8. Note that /(x^) = -p^r = T77 = 1' ^^^
\8/4\ 8/4
Ifixs) - L| = |1 - L|. Because -L > 0,1 - L > 1. Therefore,
|/(X5)-L| = |l-L| = l-L>l>f.