Proving that f'(0) exists

[Demon117]

I wanted to see what kind of responses I would get regarding this problem:

Let f : \Re\rightarrow\Re be a continuous function that is differentiable for all nonzero x such that f '(x) exists. If f'(x) \rightarrow L as x\rightarrow0 exists, prove that f '(0) exists.

[JG89]

Use the Mean Value Theorem

[Fredrik]

A few LaTeX tips: Put tex or itex tags around the whole formula instead of around each symbol, and use \mathbb R for the set of real numbers. Example: f:\mathbb R\rightarrow\mathbb R. (Click the quote button to see what I did).

[HallsofIvy]

While derivatives are not necessarily continuous, they do satisfy the "intermediate value property" (if f'(a)< c< f'(b), then f'(d)= c for some d between a and b). You can show that using the mean value theorem JG89 suggests. From that it follows that if \displaytype\lim_{x\to a}f'(x) exists, then so does f'(a) and \displaytype f'(a)= \lim_{x\to a}f'(x)