If $f:(a,b)\to \mathbb{R}$ is differentiable in a point $x\in (a,b)$, show that $\lim\limits_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}$ exists 
If $f:(a,b)\to \mathbb{R}$ is differentiable in a point $x\in (a,b)$,
  show that $\lim\limits_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}$ exists and
  that it is equal to $f'(x)$.

I figured that since the change in $x$ is described trough $2h$ that there is a $\delta$-neighbourhood around $x$ such that $|x-h|=|x+h|$. Sadly, I don't know how to make use of that argument in my proof. I know for a fact that the quotient describes to neighborhoods that, when $h\to 0$, squeeze the point.

Hint:
Just consider
$$\frac{f(x+h)-f(x-h)}{2h} =\frac{1}{2}\left(\frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h}\right) $$
