Let be $f:\mathbb{R}\to\mathbb{R}$ a differentiable function and $g:\mathbb{R}\to\mathbb{R}$ with $g(x):=\begin{cases}f(x),&x\in~\!\!]-\pi,\pi]\\f(\pi),&x=-\pi.\end{cases}$ its $2\pi$-periodic extension.
Show that $g$ is left- and right-sided differentiable.
The sample solution only says that $g$ is left-/right-sided differentiable because $f$ is differentiable and $g(\pi)=g(-\pi)$. If we look at $x\in~\!\!]-\pi,\pi[$ then it's clear. But if $x=-\pi$ then it is not obvious to me why $g$ should be left-sided differentiable just by mentioning the above argument.
So far, the only reason I found was applying the composition of limits which looks a bit complicated:
We know that $\lim\limits_{x\nearrow \pi}\frac{g(x)-g(\pi)}{x-\pi}=f'(\pi)$. Also it is clear that $\lim\limits_{y\nearrow-\pi}y+2\pi=\pi=\lim\limits_{x\nearrow \pi}x$. Taking the composition yields$$\lim\limits_{y\nearrow-\pi}\frac{g(y+2\pi)-g(\pi)}{y+2\pi-\pi}=\lim\limits_{y\nearrow-\pi}\frac{g(y)-g(-\pi)}{y-(-\pi)}=f'(\pi).$$
Maybe there is a more straight forward/swifter reason that proves the claim?