Equivalent Definition of the Derivative
Solution 1:
You just need $f$ to be continuous, and be differentiable at $x$. If $h \neq 0$, the $p$ is continuous at $h$ since $f$ is continuous. Cearly $p(h) \rightarrow 0$ as $h \rightarrow 0$ (because $f'(x)$ exists), then $p$ is also continuous at $0$