If f is continuous and differentiable on $\mathbb{R} \setminus \{0\}$, prove that if $\lim_{x\to 0}f'(x)=L$, then $f'(0) = L$ If $f$ is continuous and derivable, prove that if $\lim_{x\to 0}f'(x)=L$, then $f'(0) = L$
I need help proving this, thank you.
 A: The function must be defined and continuous at $0$ for this to hold.


*

*If $f$ is not defined at $0$, it cannot have a derivative at $0$.

*If $f$ is defined at $0$, but not continuous, it cannot have a derivative at $0$.


The result is now a simple application of l’Hôpital:
$$
\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{f'(x)}{1}
$$
which exists by assumption
A: Thanks for asking this. It's used as a Lemma in Spivak's book, and I've never been quite certain how to prove it until now. 
This is a consequence of Darboux's theorem that derivatives have the intermediate value property (i.e., if $f'(a) = A$ and $f'(b)= B$ and $A < C< B$, then there's a $c$ between $a$ and $b$ with $f'(c) = C$.)
For a discussion of that theorem, see: 
How to prove that derivatives have the Intermediate Value Property
To prove the claim from Darboux: 
First, let 
$$
g(x) = f(x) - Lx.
$$
Then we have that 
$$
\lim_{x \to 0} g'(x) = 0
$$
and want to prove that $g'(0) = 0$. What's given is that $g$ is everywhere differentiable on $\Bbb R$ (I assume!). In particular, $g'(0)$ exists. 
$$
\newcommand{\ep} {\epsilon}
$$
Suppose that $A = g'(0) \ne 0$. Without loss of generality, suppose then that $A > 0$ (the other case is very similar). 
Consider $\ep = A/3$. 
For some $\delta$, we know that $0 < |x| < \delta \implies  |f'(x)| < \ep$ by the limit assumption. In particular, picking $x_0 = \delta/2$, we have
$$
|f'(x_0)| < \frac{A}{3}.
$$
Darboux's theorem then tells us that for some $x_1$ between $0$ and $x_0$, we have 
$$
f'(x_1) = \frac{2A}{3}
$$
because $\frac{2A}{3}$ is between $A = f'(0)$ and $\pm \frac{A}{3}$, which is $f'(x_0)$. But this is a contradiction, for $|x_1| < |x_0| = \frac{\delta}{2} < \delta$, and the limit condition tells us that 
$$
|f'(x_1)| < \epsilon = \frac{A}{3}.
$$
But $\frac{2A}{3}$ is NOT less than $\frac{A}{3}$, because $A$ is positive. 
A: This is not true, x=0 is not necessarily on the domain of $f'(x)$. 
For example:
$$f:(0,1) \to R, \\
f(x)=\sin(x), f'(x)=\cos(x), \\
\lim_{x \to 0} f'(x)=1$$
but nor $f(0)$ neither $f'(0)$ exist, because they are outside of the domain, which exclude the point $x=0$. 
if $f$ is differenciable, then by definition, the derivative exist on its domain, the value $f'(0)$ exist, and in particular, the limit $\lim_{x\to o}f'(x)$ exist.
A: You want to prove that $\lim_{x\to0}\frac{f(x)-f(0)}x=L$. Take $\varepsilon>0$. Since $\lim_{x\to0}f'(x)=L$, there is a $\delta>0$ such that $|x|<\delta\Longrightarrow|f'(x)-L|<\varepsilon$. If $|x|<\delta$, take $c\in(0,x)$ such that $\frac{f(x)-f(0)}x=f'(c)$; such a $c$ exists by the mean value theorem. So,$$\left|\frac{f(x)-f(0)}x-L\right|=\bigl|f'(c)-L\bigr|<\varepsilon.$$
