# Proving that f'(0) exists

1. Sep 27, 2010

### 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.

2. Sep 27, 2010

### JG89

Use the Mean Value Theorem

3. Sep 28, 2010

### Fredrik

Staff Emeritus
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).

4. Sep 28, 2010

### 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 $\displaytype f'(a)= \lim_{x\to a}f'(x)$