True or False:
Let for . Then , so is one-one, mapping to itself.
Define , . The given function is , where the limit exists, and .
It follows that , and so .
Therefore , which implies that on .
As for the limit, we know that for , and so for .
Given we may choose depending on such that for all .
It follows that for all .
This means that in this interval. Also if maps to itself, since is closer to than is. Thus
. It follows that for all in the interval .
Since is arbitrary (get this!) we have for all .
If this works I will definitely LOL.