I need some help here.
Let $f:]a,b[$ and $c\in ]a,b[$ be such that $f$ is continuous in $c$ and $f'(x)$ exists for each $x\in ]a,c[\cup]c,b[$. If $\lim_{x \to c}f'(x)$ exists, then prove that $f'(c)$ exists and is equal to this limit.
So far, what im thinking of is to prove that f'(x) is continuous in c, that shall do it. But the problem is evedently that "$f'(x)$ exists for each $x\in ]a,c[\cup]c,b[$", thus $f'(c)$ does not have to exist... I'm stuck.
Let us choose $\delta \gt 0$ such that $(c,c+\delta)\subset (a,b)$
$f$ is continous on $[c,c+\delta]$ and $f$ is differentiable on $(c,c+\delta)$
Let $\{h_n\}$ be a sequence such that $h_n \gt 0$ and $h_n \to 0$ as $ n\to \infty$
So $, \exists k $, such that $0\lt h_n \lt \delta, \forall n\gt k$
By Mean Value Theorem on $[c,c+h_n]$ for $n\ge k$
$\frac{f(c+h_n)-f(c)}{h_n}=f'(c+\theta_n h_n)$
where $ 0\lt \theta_n \lt 1$
We have , $\lim_{n\to \infty}\theta_n h_n=0$, since $\theta_n$ is bounded.
Again $\lim_{x\to c}f'(x)=l $
$\Rightarrow \lim_{x\to c+}f'(x)=l$
By sequential criterion ,
$\lim_{n\to \infty}f'(c+\theta_n h_n)=l$
It follows that $\displaystyle\lim_{n\to \infty} \frac {f(c+h_n)-f(c)}{h_n}=l$
Since $\{h_n\}$ is an arbitary sequence converging to $0$ ,
$\displaystyle\lim_{h\to 0+}\frac{f(c+h)-f(c)}{h}=l.$
That is $Rf'(c)=l$
Similarly show that $Lf'(c)=l$. Hence $f'(c)=l$
By $Rf'(c)$ , I mean the right hand derivative and similarly for left hand derivative.
Hint:
Use the Mean value theorem and some topology to show the limit of the rate of variation $\;\dfrac{f(x)-f(c)}{x-c}$ exists and is the same as the limit of the derivative when $x$ tends to $c$.
