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.