Prove that if $f$ is differentiable at $c$ (i.e., $\lim_{x\to c} {f(x)-f(c)\over x-c}$ exists), then $f'(c) = \lim_{h\to 0}{f(c+h)-f(c)\over h}.$ 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:
$\lim_{x\to a} {f(x)-f(a)\over x-a}$.
 A: 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.
A: 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.
A: 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:

*

*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 $df(x)(h)=Ah$


*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.
