$f$ differentiable except in $1$ point Assume we have a function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that  $f$ is differentiable for all non-zero $x$, $f$ is continuous at $0$ and 
$$\lim_{x \uparrow 0} f'(x) = \lim_{x \downarrow 0} f'(x) < \infty. $$ 
Is it then true that $f$ is also differentiable at $0$?
If we drop the hypothesis that $f$ is continuous, this is not true since the signums function is a simple counterexample. But what if $f$ is continuous? I don't think this is true but I can't find a counterexample...

Yes, $f$ is differentiable at $0$.
Hint. By the Mean Value Theorem used in the interval $[0,x]$ for all $x>0$,
$$\frac{f(x)-f(0)}{x-0}=f'(t_x)$$
for some $t_x\in (0,x).$ 
Hence 
$$\lim_{x\to 0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^+}f'(t_x)=\lim_{t\to 0^+}f'(t)$$
The same can be done in $[x,0]$ fo $x<0$.
