Q:

If $f$ is a differentiable function such that $f(x)=f(-x)$, how can I prove that $f'(0)=0$ using solely the limits definition of differentiability?

Struggling to get to grips with how to go about this question: 
"Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that $f(x) = f(−x)$ for all $x$. Show, just using the definition of differentiability, that
$f'(0) = 0$."
It is obvious that $f(x)$ cannot be below $0$ given the conditions above, the function will be symmetric (mirrored in $y$-axis).
We know that $f$ is differentiable and therefore that:
$$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
Would the proof be as simple as substituting $x = 0$ and making $f(x+h) = f(0)$ as $h$ tends to $0$ ?
I appear to be missing some crucial steps here.

A:

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
As $f$ is differentiable, put $x=0$,
$$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}$$
Let $k=-h$. Then,
$$f'(0)=\lim_{k\to0}\frac{f(-k)-f(0)}{-k}$$
$$f'(0)=-\lim_{k\to0}\frac{f(k)-f(0)}{k}=-f'(0)$$
Thus,
$$f'(0)=0$$

