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 ?
Best Answer
Figured out that applying De L'Hospital provides a brutally fast proof (assuming $f$ is continuous at $x_0$) :
$$\lim_{x \to x_0}\frac{f(x) - f(x_0)}{x-x_0} = \lim_{x \to x_0} \frac{[f(x) - f(x_0)]'}{(x-x_0)'}=\lim_{x\to x_0} \frac{f'(x)}{1}=c \implies f'(x_0) = c$$