Proof of Fermat Stationary Point Criterion

Let $f$ be differentiable at $x_0$ and suppose $x_0$ is a local extremum. If $x_0$ is a local maximum, then for $h>0$ small, $\frac{f(x_0+h)-f(x_0)}h\le0,$ while for $h<0$ small the same inequality implies $\frac{f(x_0+h)-f(x_0)}h\ge0.$ Taking $h\to0^+$ and $h\to0^-$ gives right derivative $\le0$ and left derivative $\ge0$. Since $f'(x_0)$ exists, both one-sided limits are equal, so $f'(x_0)=0$. The local minimum case is identical with inequalities reversed.