Suppose $f\in C(\mathbb R)\cap C^1(\mathbb R\setminus \{ 0\})$ and $\lim_{x\to 0}f'(x) $ exists and equals $L\in\mathbb R$ . Does this imply that $f\in C^1(\mathbb R)$?
Derivatives cannot have jump singularities, since they satisfy the intermediate value property. Therefore, if $f'(0)$ exists, its value must be $L$, and $f$ is then $C^1$. But this doesn't prove the result.
I also can't imagine a counterexample...
Remarks:
- This seems simple enough that it should have been asked before, but I can't find a duplicate. Moreover, (see rem. 3) its not so unnatural to assume its true. This is kind of close. Most questions with similar names seem to be about multivariate calculus.
- Note the standard example of a function whose derivative is not continuous at 0 does not help ($f(x) = x\sin(1/x)$)
- If true, this result would validate attempts to prove the differentiability of e.g. $e^{-1/|x|}$ by establishing the existence of the above limit instead.
By l'Hopital's rule you have that $$ \lim_{x\to 0} \frac{f(x) - f(0)}{x} = \lim_{x\to 0} f'(x) = L, $$ hence $f$ is differentiable at $0$ and $f'(0) = L$. We can then conclude that $f\in C^1(\mathbb{R})$.