sufficient condition for derivative to be continuous? let $f:\mathbb{R} \rightarrow\mathbb{R}$ be a continuous function in $x_0\in\mathbb{R}$.

In addition, $f'(x)$ is defined for every $x\in(x_0-\delta,x_0+\delta),x_0\notin(x_0-\delta,x_0+\delta)$

In addition, i know that $\underset{x_0}{lim}    f'(x)=L$

is it true that $f'(x)$ is continuous  at every  $x\in(x_0-\delta,x_0+\delta),x_0\notin(x_0-\delta,x_0+\delta)$
Assuming I understand your notation, no. 

Let 
$$
g(x) = \begin{cases} x^2 \sin \frac{1}{x} &amp; x \ne 0 \\ 0 &amp; x = 0\end{cases} 
$$

$g$ is everywhere differentiable, but the derivative is not continuous. 

Now let 
$$
f(x) = g(x - \frac{\delta}{2}) + g(x + \frac{\delta}{2})
$$
It's everywhere differentiable, but the deriv is not continuous at the points $\pm \delta/2$.
$f'(x)$ must not be continuous at every $x \in (x_0 - \delta, x_0 + \delta)$. 

Example: May be $rect_\delta(x-x_0)$ The rectangle function with Center at Point $x_0$ and broadness $\delta$. Then $f'(x)=rect_\delta(x-x_0)$ is discontinuous at $x_0= \delta/2$ and $x_0=- \delta/2$. The function $f(x)$, however is continuous.