Intro analysis question related to MVT 
Q) Let $f$ be a real-valued function which is continuous in a
  neighbourhood $N$ of some point $c∈R$. Suppose that $f$ is
  differentiable on N\ {c} and that $\lim_{x->c} f'(x)=L$. Show that $f$
  is differentiable at $c$ with $f'(c) = L$.

My try: $f$ is differentiable at $c$ if the limit $\lim_{x->c+}(f(x)-f(c))/(x-c)$ exists. As $f$ is differentiable in $(c, c+h)$, from MVT, there is some $y\in (c, c+h)$ such that interval such that $f'(y)=\frac{f(c+h)-f(c)}{c+h-c}=L$. This implies that: $\lim_{x->c} f'(y)=\lim_{x->c} L = f'(c)=L$. 
I know this isn't correct, but I don't know where I'm going wrong.
 A: $f'(y)=L$ is not correct. 
Let $\epsilon >0$ and choose $\delta >0$ such that $|f'(y)-L| <\epsilon$ for $|x-c| < \delta$. Then for suitable $y$ we get  $|\frac {f(c+h)-f(c)} h -L|=|f'(y)-L|<\epsilon$ for $|x-c| < \delta$.
A: 1) Let $x >c$.
$\dfrac{f(x)-f(c)}{x-c}=f'(t)$ , $c <t< x$.
$\lim_{x \rightarrow c^+} t= c$, hence
$\lim_{x \rightarrow c^+} \dfrac{f(x)-f(c)}{x-c}=$
$\lim_{t \rightarrow c^+}f'(t)=L$;
2) Likewise for $x <0$.
3) 1)and 2) imply $f$ differentiable at $c$, and $f'(c)=L$.
A bit formal:
0) $x >c$;
1) Need to show :
For $\epsilon >0$ given,
there is a $\delta$ s.t. 
$0<x-c<\delta$ implies $|\dfrac{f(x)-f(c)}{x-c}-L|<\epsilon$.
2) We have
$\lim_{ t \rightarrow c^+}f'(t)=L$, i.e.
there is a $\delta_1$ s.t. 
$0<t-c<\delta_1$ implies $|f'(t)-L|<\epsilon.$
Let $\delta=\delta_1$; then for
$0<x-c < \delta$ we have
$|\dfrac{f(x)-f(c)}{x-c}-L|=$
$|f'(t)-L|<\epsilon$,
since $\dfrac{f(x)-f(c)}{x-c} = f'(t)$ , $0<t-c<x-c$.
3) Consider $x \rightarrow c^-$  and complete the proof.
A: Alternatively, you can use de l'Hospital rule, whose hypotheses are satisfied in the right neighboorhood $N^+$ and left neighboorhood $N^-$ of $c$, so that
$$\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c}\stackrel{H}{=}\lim_{x\to c^+}f'(x)=L,$$
and 
$$\lim_{x\to c^-}\frac{f(x)-f(c)}{x-c}\stackrel{H}{=}\lim_{x\to c^-}f'(x)=L.$$
Hence, the thesis.
A: Using sequences is quite comfortable here.
Choose in $N$ any sequence $x_n \stackrel{n\to \infty}{\rightarrow} c$ ($x_n \neq c$ for all $n$)
MVT tells you, that for each $x_n$ there is $\xi_n$ with
$$\frac{f(x_n) - f(c)}{x_n -c}= f'(\xi_n) \mbox{ s.t.  } 0 < |\xi_n - c| < |x_n - c|$$
If follows $\xi_n \stackrel{n\to \infty}{\rightarrow} c$ and 
$$\lim_{n\to \infty}\frac{f(x_n) - f(c)}{x_n -c} = \lim_{n\to \infty} f'(\xi_n) =L$$.
