Prove a derivative exists $f\colon (a,b) \to \mathbb{R}$ is continuous, with finite derivatives everywhere in $(a,b)$, except maybe at $c$.
If 
$$\lim_{x\to c}f'(x) = B,$$
show that $f'(c)$ exists and equals $B$.
I'm not sure where to start on this. I've tried using the definitions of continuity and f' but it isn't working out  
Well I started out with 
$$\lim_{x\to c}\frac{f(x)-f(c)}{x-c} = f'(c)$$
is equivalent to saying
given $\epsilon\gt 0$ there exists $\delta\gt 0$ such that
$$|x-c| \lt \delta \Longrightarrow \left|\frac{f(x)-f(c)}{x-c} - f'(c)\right| \lt\epsilon.$$  
I was trying to prove $f'(c)$ must exist since $f$ is continuous but I feel like I'm assuming what I'm trying to prove. 
 A: You are right that you are kind of assuming what you want, because you cannot even write down $f'(c)$ until you prove it exists; this can be fixed if you replace $f'(c)$ by a specific limit, and the specific limit you want is probably $B$. But even so ran into a bit of an alley, so let me help you out.
Here is the beginning of half a proof, following the Hint of Robin Chapman:
Consider first the limit as $x\to c^+$. For any $h\gt 0$, the function $f(x)$ is continuous on $[c,c+h]$, and is differentiable on $(c,c+h)$, so by the Mean Value Theorem there exists a point $d_h$ (which depends on $h$) such that 
$$\frac{f(c+h)-f(c)}{h} = f'(d_h).$$
Therefore, we have that
\begin{align*}
\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c} &= \lim_{h\to 0^+}\frac{f(c+h)-f(c)}{h}\
&=\lim_{h\to 0^+}f'(d_h).
\end{align*}
So we've managed to turn the (one-sided) limit that will tell us whether $f$ is differentiable at $c$ into a limit about $f'(x)$. Luckily, we have information about limits of $f'(x)$. So perhaps we can leverage that information into an actual value for this limit? If so, perhaps we can try doing the same thing for the limit from the left (as $x\to c^-$), so that the two together shows us that 
$$\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$
exists; and will tell us what the value of $f'(c)$ must be in that case.
A: So what can go wrong?  We can certainly define a function g(x)=something nice except when x=c, something different at c.  Then $\lim_{x\rightarrow c} g(x) \neq g(c)$ Can you rule out this behavior in $f'(x)$?  Or maybe $f'(x)$ has a step at $c$.  Why can't it?
