Suppose $f:[a,b]\rightarrow \mathbb{R}$ is a continuous function that is differentiable on the open interval $(a,c) \cup (c,b)$ for some $c\in [a,b]$. Show that if $\lim_{x\to c} f'(x)$ exists, then $f'(c)$ exists and $f'$ is continuous at $c$.
I'm a measly undergrad so I have very little intuition about this problem. My main point of confusion is this: it seems that even for the points where we know $f'$ exists, there is no guarantee that $f'$ is continuous. So at $c$, we only know that the limit of the derivative exists. To me, this feels like a weaker condition than the derivative itself existing - why should we be able to expect that it is continuous?
Hint: Let $\lim_{x\rightarrow c} f'(x) = L$. You want to show \begin{align} \lim_{x\rightarrow c} \frac{f(x)-f(c)}{x-c} \end{align} exists and is equal to $L$. By the mean value theorem, we see that \begin{align} \left|\frac{f(x)-f(c)}{x-c}-L\right| = \left|f'(y)-L \right| \end{align} for $x<y<c$ (likewise $x>y>c$ if you are considering the other side limit). If $x$ is close to $c$ then $|f'(y)-L|$ is small (why?). This should prove that $f$ is differentiable at $c$. Using Darboux's Theorem you should be able to show that $f'$ is continuous.