This seems like another good question for consideration. I think the answer is yes just because I cannot think of a way to make it break down because the domain is defined for all of $\mathbb{R}$. Any thoughts?
Thank you in advance.
This question is not a duplicate, it deals with the limit of the derivative specifically at 0. The other two cited questions do not.
Thank you.
Yes, you actually can conclude that. If $h$ is a nonzero real number, then by the mean value theorem there exists $x_h\in\mathbb{R}\backslash\{0\}$ with $|x_h|<|h|$ and $\frac{f(h)-f(0)}{h-0}=f'(x_h)$. With the condition we then obtain $$ f'(0)=\lim_{\substack{h\to0 \\ \neq}}\frac{f(h)-f(0)}{h-0}=\lim_{\substack{h\to0 \\ \neq}}f'(x_h)=L. $$ Note that this is quite remarkable; many comments pointed out that this cannot be true because then all derivatives would be continuous. However, it shows the weaker but nevertheless interesting property, that if the limit exists, then it is continuous. Also note that a priori we didn't even require $f$ to be differentiable in $0$ (although it is, but a posteriori).
Edit:
As pointed out in the comments, there are doubts of $\lim_{\substack{h\to0 \\ \neq}}f'(x_h)=L$ being true. Here is a precise proof:
Let $\varepsilon>0$, then by the assumption $\lim_{\substack{h\to0 \\ \neq}}f'(h)=L$, there exists $\delta>0$ such that for all $h\in(-\delta,\delta)\backslash\{0\}$ we have $$ |f'(h)-L|<\varepsilon. $$ But if $h\in(-\delta,\delta)\backslash\{0\}$ then with $|x_h|<|h|$ we have also $x_h\in(-\delta,\delta)\backslash\{0\}$. Therefore we conclude $$ |f'(x_h)-L|<\varepsilon\qquad\forall h\in(-\delta,\delta)\backslash\{0\}. $$ As $\varepsilon>0$ was chosen arbitrarily, we hence obtain $$ \lim_{\substack{h\to0 \\ \neq}}f'(x_h)=L. $$
Yes, you can, even with weaker assumptions on $f$.
Suppose $f$ is continuous on $(a,b)$ and that it is differentiable on $(a,b)$ except perhaps at $c\in(a,b)$. If $\lim\limits_{x\to c}f'(x)=L$, then $f$ is differentiable also at $c$ and $f'(c)=L$.
Indeed, the functions $f(x)-f(c)$ and $x-c$ satisfy the hypotheses of l’Hôpital’s theorem at $c$, so $$ \lim_{x\to c}\frac{f(x)-f(c)}{x-c}= \lim_{x\to c}\frac{f'(x)}{1}=L $$ Note that continuity of $f$ at $c$ is crucial for being able to apply l’Hôpital. No assumption is made on the continuity of $f'$.
Note also that $f$ might be differentiable at $c$ even if the limit at $c$ of the derivative doesn't exist. The standard example is $f(x)=x^2\sin(1/x)$ for $x\ne0$ and $f(0)=0$.
Theorem
Suppose that $f$ is continuous on $\mathbb{R}$ and differentiable except perhaps at $a\in \mathbb{R}$. Suppose further that $\lim_{x\rightarrow a}f'(x)= L<\infty$ exists. Then $f$ is differentiable at $a$ with derivative $f'(a)=\lim_{x\rightarrow a}f'(x)$.
Proof: Let $h>0$. $f$ is continuous on $[a,a+h]$ and differentiable on $ (a,a+h)$. Then, by the Mean Value Theorem, $\exists\,c_h\in(a,a+h)$ such that
$$ f'(c_h)=\frac{f(a+h)-f(a)}{h}.$$
Now taking the limit as $ h\rightarrow 0$ on both sides:
$$\lim_{h\rightarrow 0^+}f'(c_h)=\lim_{h\rightarrow 0^+}\frac{f(a+h)-f(a)}{h}.$$
Now $c_h\rightarrow a^+$ as $h\rightarrow 0^+$ hence the left hand side is equal to $L$:
$$ \lim_{h\rightarrow 0^+}\frac{f(a+h)-f(a)}{h}=L.$$
Similarly taking $ h<0$ and $ h\rightarrow 0^-$ we can show:
$$ \lim_{h\rightarrow 0^-}\frac{f(a+h)-f(a)}{h}=L.$$
Hence $ f$ is differentiable at $a$ with derivative $f'(a)=L$ $\bullet$
