If f'(x) → L as x → 0, does f'(0) exist?

[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$$\rightarrow$$0 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 $\displaytype f'(a)= \lim_{x\to a}f'(x)$