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.
Nevertheless, that answer is done and accepted, but looking comments under OP, let me say, that here we have question about equivalence of **differentiability** and **existence of derivative**:

1. **Differentiability** (Rudin W. - Principles of mathematical analysis-(1976) 212-213 p.). Suppose $f$ is defined on $(a,b) \subset \mathbb{R}$. We say, that $f$ is differentiable in $x \in (a,b)$ if exists linear mapping $A:\mathbb{R} \to \mathbb{R}$, such that
$$\lim\limits\_{h \to 0}\frac{f(x+h)-f(x)-Ah}{h}=0$$
this linear mapping we call differential and denote $df(x)(h)=Ah$
2. **Derivative**.(Same book as above 211p) We say, that $f$ have derivative in $x \in (a,b)$, when exists
$$\lim\limits\_{h \to 0}\frac{f(x+h)-f(x)}{h}=B$$
and $B$ we denote as $B=f'(x)$

Now first sentence of OP mean, that this 2 definitions are equivalent: if exists $A$ in first, then exists $B$ from second and they are equal. Reverse also.

So $df(x)(h)=f'(x)h$. More [here](https://math.stackexchange.com/questions/3819116/rigorously-whats-happening-when-i-treat-fracdydx-as-a-fraction/3819142#3819142).