Question: Prove that if $f$ is differentiable at $c$, then
$f'(c) = \lim\_{h\to 0}{f(c+h)-f(c)\over h}$.

I did this: If f is differentiable at $c$, then it's continuous at $c$, which implies,

$\lim\_{x\to c}$ $f(x)=f(c)$ if only if $\lim\_{h\to 0}$ $f(c+h)-f(c)=0$.

Then dividing by $h$,

$$\lim\_{h\to 0}{f(c+h)-f(c)\over h}=f'(c).$$

I don't know if this is correct, I feel it's kind of arbitrary.

Edit: The definition of derivative is as follows-
If $f$ is differentiable at a point *a* then the following limit exists:
$\lim\_{x\to a} {f(x)-f(a)\over x-a}$.

Answer: 
So, from what I understand, the question boils down to showing the equivalence of two definitions of the derivative. Here's what you begin with:

> 
> If $f$ is differentiable at $x=a$, the following limit exists and is equal to $f'(a)$:
> $\lim\_{x\to a} {f(x)-f(a)\over x-a}$
> 
> 
> 

Now, put $x-a=h$, that is $x = a + h$. $x \to a$ is the same as $h \to 0$, so you finally get:

$f'(a) = \lim\_{h\to0} {f(a+h)-f(a)\over h}$ as desired.
The definition of $f'(c)$ is that if $\lim\_{h\to 0}\frac {f(c+h)-f(c)}{h}$ exists then this limit is $f'(c),$ and if this limit doesn't exist then $f'(c)$ doesn't exist.

"$f $ is differentiable at $c$" is just another way to say that $f'(c)$ exists.