Can only the existence of the right and left derivatives imply continuity? If $f:X\rightarrow \mathbb{R} \,$ is a function with $x_0 \in \overline{X}   \,\setminus \partial(\overline{X}) $ such that :
$$\exists \,\,\,\,f'_-(x_0)=\lim_{x\rightarrow x_0^-}\dfrac{f(x)-f(x_0)}{x-x_0},$$
$$\exists \,\,\,\,f'_+(x_0)=\lim_{x\rightarrow x_0^+}\dfrac{f(x)-f(x_0)}{x-x_0}$$
 but with possibly $f'_-(x_0) \not= f'_+(x_0)$,  does this still imply continuity of $f$ ? 

Suppose that$$\lim_{x\to{x_0}^+}\frac{f(x)-f(x_0)}{x-x_0}$$exists (in $\mathbb R$). Then\begin{align}\lim_{x\to{x_0}^+}f(x)&=f(x_0)+\lim_{x\to{x_0}^+}f(x)-f(x_0)\\&=f(x_0)+\lim_{x\to{x_0}^+}\left((x-x_0)\frac{f(x)-f(x_0)}{x-x_0}\right)\\&=f(x_0)+\lim_{x\to{x_0}^+}(x-x_0)\lim_{x\to{x_0}^+}\frac{f(x)-f(x_0)}{x-x_0}\\&=f(x_0)+0\\&=f(x_0).\end{align}For the same reason, if$$\lim_{x\to{x_0}^-}\frac{f(x)-f(x_0)}{x-x_0}$$exists (in $\mathbb R$), then $\lim_{x\to{x_0}^-}f(x)=f(x_0)$.
