Is $\lim\limits_{x\to x_0}f'(x)=f'(x_0)$? Let $f$ be a function defined in the open interval $(a,b)$ and let $x_0\in(a,b)$. Suppose in addition that $f'(x)$ exists for all $x_0\neq x\in(a,b)$. Is the following statement true: 

If $\lim\limits_{x\to x_0}f'(x)$ exists, then $f'(x_0)$ exists and $\lim\limits_{x\to x_0}f'(x)=f'(x_0)$.

Thanks! 
 A: A qualified "yes": If $f$ is continuous at $x_{0}$, and if $\lim\limits_{x \to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.
Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)
The claim follows from the Mean Value Theorem: If $\delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < \delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that
$$
\frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t).
$$
Since $|t - x_{0}| < |h|$, taking the limit as $h \to 0$ forces $t - x_{0} \to 0$, as well, so
$$
f'(x_{0}) = \lim_{h \to 0} \frac{f(x_{0} + h) - f(x_{0})}{h}
  = \lim_{t \to x_{0}} f'(t).
$$
(Continuity of $f$ was needed to invoke the Mean Value Theorem.)
A: An easier proof, for the chosen answer, is by using the L'Hopital's rule.
We know that $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$. Now if f is continuous at $a$ the we have a $\frac{0}{0}$ situation, and we can apply the  L'Hopital's rule to see that if the limit of $f(x)$ when $x\mapsto a$ exists then it is equal to $f'(a)$.
A: No, it is not true. For a simple counterexample take the indicator function $\chi_{[0,1)}$ in $x\in(-1,1)$ and take $x_0$ to be 0.
A: Not neccessarily true. We need $f'$ to be continuous at $x_0$. Actually the conditions that $\lim_{x\to x_0} f'(x)$ and $f'(x_0)$ exist and they are equal to each other are the conditions for the function $f'$ to be continous.
Here's a counter-example: Consider the function:
$$f(x) = \begin{cases} 0, & \mbox{if } \mbox{$x \le 0$} \\ x, & \mbox{if } \mbox{$x>0$} \end{cases}$$
Obviously $f$ is defined on $\mathbb{R}$, but the function isn't differentiable at $0$ and $f'$ is discontinuous at $0$. So by choosing $x_0 = 0$ all the conditions are satisfied, but the result isn't, hence the implication is wrong.
