A question about existence of derivative of function at Zero Assume that $f:\mathbb{R}\to\mathbb{R}$ is continuous and differentiable everywhere but at $0$.
If $\displaystyle\lim_{x\to0} f'(x) = L$ exists, then does it follow that $f'(0)$ exists?
Prove or disprove.
I think it has to be true. I know that by definition $\displaystyle f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}$, but I could not able to further steps from here.
 could you please help me out.
 A: By the mean value theorem there is $c_h\in(0,h)$ such that
$$\frac{f(h)-f(0)}h=f'(c_h)$$
so pass to the limit $h\to0^+$ and you get $f_r'(0)=L$. Similarly you get $f'_l(0)=L$. Conclude.
A: This is an application of L'Hospital's Rule:
$$f'(0) = \lim_{t \to 0} \frac{f(t)-f(0)}{t-0} = \lim_{t \to 0} f'(t)=L$$ 
where the second equality is given by the rule.
A: There is an ambiguity in your question. I assume you mean $f(x)$ is continuous everywhere but at $x=0$ and differentiable everywhere but at $x=0$. In this case what you said and the converse of what you said can be violated.
I give you some examples. Consider the following function
$$f(x) = \left\{ \matrix{
  {x^2}\sin ({1 \over x})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ne 0 \hfill \cr 
  0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \hfill \cr}  \right.$$
It's derivative is
$$f'(x) = \left\{ \matrix{
  2x\sin ({1 \over x})\, - \cos ({1 \over x})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ne 0 \hfill \cr 
  0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \hfill \cr}  \right.$$
You can simply see that $f'(0)$ exist since
$$f'(0) = \mathop {\lim }\limits_{x \to 0} {{f(x) - f(0)} \over {x - 0}} = \mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = \mathop {\lim }\limits_{x \to 0} {{{x^2}\sin ({1 \over x})} \over x} = \mathop {\lim }\limits_{x \to 0} x\sin ({1 \over x}) = 0$$
but $\mathop {\lim f'(x)}\limits_{x \to 0} $ doesn't exist. Hence, the existence of $f'(0)$ doesn't imply the existence of $\mathop {\lim f'(x)}\limits_{x \to 0} $.
The vice versa can also happen, i.e., $\mathop {\lim f'(x)}\limits_{x \to 0} $ exists but $f'(0)$ doesn't exist. For this case you can consider the following simple function
$$f(x) = \left\{ \matrix{
  \sin (x)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ne 0 \hfill \cr 
  1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \hfill \cr}  \right.$$
which is a discontinuous function at $x=0$ and hence $f'(0)$ doesn't exist but simply you can check that $\mathop {\lim f'(x)}\limits_{x \to 0} $ exists.
A: If you take:
$$\begin{align}
f(x) = \begin{cases}
         x^2 & x \ne 0 \\
         1   & x = 0
       \end{cases}
\end{align}$$
then $f'(x) = 2 x$ unless $x = 0$, and $\lim_{x \to 0} f'(x) = 0$. But the definition of the derivative at $x = 0$ yields:
$$
\lim_{h \to 0} \frac{f(h) - f(0)}{h} = \frac{h^2 - 1}{h}
$$
This just doesn't exist. The derivative has a "hole" (more precisely, a removable singularity) at $x = 0$.
