Cfrgtkky

Suppose that $lim limits_{x to c} f'(x) = l in Bbb
R$.

1) Does it mean that $f$ is differentiable at $c$ and $f'(c) = l$?

2) Is it possible that $f'(c)$ does not exist?

3) What is the difference whether $f$ is continuous or not?

4) If $f$ is continuous, how to prove the statement?

Yes, if $lim limits_{x to c} f'(x) = l in Bbb
R$ then $f$ is differentiable at $c$ provided that $f$ is continuous at $c$. Then by the Mean Value Theorem, there is $t_x$ between $x$ and $c$ such that
$$frac{f(x)-f(c)}{x-c}=f'(t_x).$$
It follows that, as $xto c$, we have $t_xto c$ and $f'(t_x)to l$.
Hence $f'(c)$ exists and it is equal to $l$.

Of course, if $f$ is not continuous at $c$ or if $f$ is not even defined at $c$, then $f$ is not differentiable at $c$. Take for example
$$f(x) = begin{cases}1 & text{if $x=0$}\ x & text{otherwise} end{cases}.$$
Then $lim_{xto 0}f'(x)=1$, but the $f$ is not differentiable at $0$:
$$lim_{xto 0^+}frac{f(x)-f(0)}{x-0}=lim_{xto 0^+}frac{x-1}{x}=-infty.$$

It is possible that $f'(c)$ does not even exist. The existence of a limit of derivative does not even guarantee continuity of $f$ at $c$. For example consider a constant function with a jump discontinuity at $c$.
$$F(x) = begin{cases}0 & x neq 1 \ 1 & x=1end{cases}$$ is not differentiable at 1 even though the limit of derivative exist when $x$ tends to 1.