If $f$ is differentiable for $x\neq x_0$ and $\lim_{x\to x_0} f'(x) = c$ then $f'(x_0) = c$ Exercise :

Let $f : \mathbb R \to \mathbb R$ and $x_0 \in \mathbb R$. Suppose that $f$ is differentiable for all $x \neq x_0$. If $\lim_{x \to x_0}f'(x) = c \in \mathbb R$ show that $f$ is differentiable at $x_0$ and $f'(x_0) = c$.

Attempt :
Isn't it pretty straight forward that since $\lim_{x \to x_0} f'(x) = c$ then $f'$ is continuous at $x_0$ and thus differentiablewith $f'(x_0) = c$ ? Does it need some more delicate or rigorous mathematical proof ?
 A: Figured out that applying De L'Hospital provides a brutally fast proof (assuming $f$ is continuous at $x_0$) :
$$\lim_{x \to x_0}\frac{f(x) - f(x_0)}{x-x_0} = \lim_{x \to x_0} \frac{[f(x) - f(x_0)]'}{(x-x_0)'}=\lim_{x\to x_0} \frac{f'(x)}{1}=c \implies f'(x_0) = c$$
A: Assuming $f$ is continuous at $x_0.$ It is equivalent  to show  $g'(0)=c$ where $g(x)=f(x+x_0)-f(x_0).$  That is, to show  $\lim_{x\to 0}\frac {g(x)}{x}=c.$
We have  $\lim_{x\to 0}g'(x)=c$ and $\lim_{y\to 0}g(y)=g(0)=0.$  
Since $\lim_{x\to 0}g'(x)=c,$ there exists $x_1>0$ such that  $g'$ is bounded on $(0,x_1).$ For $x\in (0,x_1)$ let $M(x)=\sup_{z\in (0,x)}g'(z)$ and $m(x)=\inf_{z\in (0,x)}g'(z).$
For $0<y<x<x_1$ we have, by the MVT, $g(x)=g(y)+(x-y)g'(z_{x,y})$ for some $z_{x,y}\in (y,x).$ So $$\frac  {g(x)}{x}=\frac {g(y)}{x}+(1-\frac {y}{x})g'(z_{x,y}).$$ Therefore  $$\frac {g(y)}{x}+(1-\frac {y}{x})M(x)\geq \frac {g(x)}{x}\geq \frac {g(y)}{x}+(1-\frac {y}{x})m(x).$$
Now keeping $x$ fixed and letting $y\to 0,$ since $\lim_{y\to 0}g(y)=0,$ we have $$M(x)\geq \frac {g(x)}{x}\geq m(x)$$ for all $x\in (0,x_1)$.
Since $c=\lim_{x\to 0+}M(x)=\lim_{x\to 0+}m(x),$ we have therefore $ \lim_{x\to 0+}\frac {g(x)}{x}=c.$
Similarly, $\lim_{x\to 0-}\frac {g(x)}{x}=c.$
Remark . The change from $f$ to $g$ is not really necessary. It reduces what I call the "clutter", to make it easier to follow.
