Why is this not correct？ We are assigned to deal with the following task.

Assume that $f(x)$ is derivable for any $x \in \mathbb{R}$. We want to
  research $\lim\limits_{x \to x_0}f'(x)$ where $x_0 \in \mathbb{R}$.

Notice that
\begin{align*}
 f'(x_0)=\lim_{x \to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x \to x_0}\frac{f'(\xi)(x-x_0)}{x-x_0}=\lim_{x \to x_0}f'(\xi),
\end{align*} 
where we applied Lagrange's Mean Value Theorem, and $ x_0 \lessgtr \xi \lessgtr x.$ Since $\xi$ is squeezed by $x_0$ and $x$, then $x \to x_0$ implies $\xi \to x_0$. Thus
$$f'(x_0)=\lim_{x \to x_0}f'(\xi)=\lim_{\xi \to x_0}f'(\xi).$$ 
What does this say? It shows that $f'(x)$ is always continuous at any point $x=x_0$, which is an absurd conclusion, because we know safely $f'(x)$ may probably has the discontinuity point (of the second kind). But where dose the mistake occur during the reasoning above?
 A: Your reasoning only shows that if $\lim_{x \to x_0} f'(x)$ exists, then it's equal to $f'(x_0)$. But, as you say, it may not exist.
This is a rather classical exercise, which has been treated many times on this site, for example in this question.
A: Actually，we can make a comment for the reasoning like this:
The fact that $\lim\limits_{\xi \to x_0}f'(\xi)$ exists dose not imply $\lim\limits_{x \to x_0}f'(x)$ also exsits, because, according to Heine's Theorem, the latter one necessitates that $f'(x_n)$ converges for any sequence $x_n \to x_0$. As we can see, $\xi_n$ is only a specific sequence. Even though $f'(\xi_n)$ is convergent, this is not enough to guarantee $f'(x_n)$ converges as well. 
A: Actually, you are right, this also called 'theorem of derivative limitation' in some textbooks.
If you already assumed that $f(x)$ is derivable for any $x \in \mathbb{R}$.
With an additional condition the limitation of derivative $\lim_{x \to x_{0}}f'(x)$ exist. 
You can write your conclusion:
$$\lim_{x \to x_{0}}f'(x)=f'(x_{0})$$
The condition $f(x)$ is derivable for any $x \in \mathbb{R}$ can also be weaken to $f(x)$ is continuous in $U(x_{0})$ which is the neighborhood of $x_{0}$, and derivable in $U°(x_{0})$ which is the punctured neighborhood of $x_{0}$.
The only problem is that you have not assumed the limitation of your derivative should always exist, which means the derivative can not divergence to infinity, like you say sometimes it would be the discontinuity point of the second kind.
And this theorem can be simply remembered as in other 'not so strict (please notice the comment below)' ways like:
If a continuous function has (finite) derivative, its derivative function is also continuous. (but maybe not derivable)
