# $\lim_{x \rightarrow 0} f'(x) = L$ exists, does it follow that $f'(0)$ exists? [duplicate]

The question is the following: Assume $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, and for all $x \neq 0$, $f'(x)$ exists. If $\lim_{x \rightarrow 0} f'(x) = L$ exists, does it follow that $f'(0)$ exists?

My intuition is that this does not have to hold in general. However, I keep finding counter examples such that $f'(0)$ exists where $\lim_{x \rightarrow 0} f'(x) = L$ does not exist, which is obviously not the original question.

## marked as duplicate by Hans Lundmark, José Carlos Santos, Sahiba Arora, Yujie Zha, Antonios-Alexandros RobotisAug 6 '17 at 13:38

Note

$$\lim_{h \to 0^{+}} \left(\frac{f(h) - f(0)}{h}\right) = \lim_{h \to 0^{+}} f'(\alpha) = L$$

where $\alpha \in (0,h)$. To be precise, the hypotheses of the mean value theorem are satisfied since $f$ is assumed continuous on $[0,h]$ and differentiable on $(0,h)$.

The above limit is necessarily $L$, since the $\alpha$ is forced to $0$ as $h \to 0^{+}$, and we know $\lim_{h \to 0} f'(h) = L$. This can easily be formalized with $\epsilon-\delta$ if so desired.

Similarly, $$\lim_{h \to 0^{-}} \left(\frac{f(h) - f(0)}{h}\right) = \lim_{h \to 0^{-}} f'(\alpha) = L$$

where $\alpha \in (h, 0)$.

This implies

$$\lim_{h \to 0} \left(\frac{f(h) - f(0)}{h}\right) = L$$

and so $f'(0)$ exists.