Show $f(x) = |x|^p$ is differentiable at $0$ for $p > 1$.
My proof is below. My question is to verify or improve it, or respond to the remark.
Proof: We examine the limit from both the right and left:
$$\begin{align} \lim_{h \to 0^+} \frac {f(x+h) - f(x)}{h} = \lim_{h \to 0^+} \frac {h^p - 0}{h} = \lim_{h \to 0^+} h^{p-1} = 0 \end{align}$$ since $p - 1 > 0$. Likewise, $$\begin{align} \lim_{h \to 0^-} \frac {f(x+h) - f(x)}{h} = \lim_{h \to 0^-} \frac {(-h)^p - 0}{-(-h)} = - \lim_{h \to 0^-} (-h)^{p-1} = 0 \end{align}.$$
Since the limits from both sides are equal, the limit is defined. QED.
Remark: At $0$, the error from the linear approximation of $f$ for a small step $h$ is $h^{p-1}$ which for $p \in (1,2)$ is only "slightly smaller" then $h$ itself, but still close enough to make the function differentiable.