Define the function:
Prove there is no such that
Proof. We prove this by contradiction. Suppose there does exist some such that
. From the definition of limit this means that for all
there exists a
such that
First, we claim that for any such we must have
. This must be the case since if
then
, so we may choose
such that
. But then, since
for all
, we have
(Where we used part (i) this exercise (Section I.4.9, Exercise #1) for the inequality .) This contradicts our choice of
, so
must be less than or equal to 1.
Next, suppose . Then choose
. To obtain our contradiction we must show that there is no
such that
By the Archimedean property of the real numbers we know that for any , there exists a positive integer
with
such that
. (The Archimedean property guarantees us an integer, but then the inequality will hold for any larger integer, so we can choose one with the property of being
.) But,
Then, from the definition of and since
and
we have,
But then,
Furthermore,
This contradicts that if then for every
we have
for all
. In other words, we found an
greater than 0 (in particular, we found
) such that no matter how small we choose
there exists an
such that
is smaller than
, but
is bigger than
. This exactly contradicts the definition of
. Hence, there can be no such number
This proof shows that there is now way to define so that
is continuous at 0. We see this since for
to be continuous at 0 we must have
. But if we try to define
for any
, we will fail to achieve continuity since the proof shows
for any
we might want to choose.