Suppose that $h$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and that $c \in (a, b)$. Suppose also that $\lim\limits_{x \to c} h'(x)$ exists. Prove that $h'$ is continuous at $c$.
I really have no idea how to think about this problem. I know if the limit exists then it's differentiable at the point, i was assuming differentiability implied continuity. Someone please give me advice.
Let $f$ be defined as
$$f(x,y)= \begin{cases} \frac{h(x+y)-h(x)}{y}&, y\ne 0\\\\ h'(x)&,y=0 \end{cases}$$
for $x\in (a,b)$ and $x+y\in (a,b)$.
Since $h$ is continuous on $[a,b]$, then for $y\ne 0$, $f(x,y)$ converges uniformly to $f(c,y)$ as $x\to c$.
In addition $\lim_{y\to 0}f(x,y)=h'(x)$.
Then, the Moore-Osgood Theorem guarantees that
$$\lim_{x\to c}\lim_{y\to 0}f(x,y)=\lim_{y\to 0}\lim_{x\to c}f(x,y)\tag 1$$
The left-hand side of $(1)$ is equal to $\lim_{x\to c}h'(x)$. The right-hand side of $(1)$ is equal to $h'(c)$.
And we are done!
This is a standard property of derivatives: derivatives can't have jump discontinuity. In other words if both $f'(c)$ and $\lim_{x \to c}f'(x)$ exist then they are equal and hence $f'$ is continuous at $c$.
To prove the above property we can use either Mean Value Theorem or Darboux Theorem. First let's use the mean value theorem. We have \begin{align} f'(c) &= \lim_{h \to 0}\frac{f(c + h) - f(c)}{h}\notag\\ &= \lim_{h \to 0}f'(c_h)\text{ (for some }c_h\text{ between }c, c + h)\notag\\ &= \lim_{x \to c}f'(x)\notag\\ \end{align}
Next we use Darboux theorem which says that derivatives possess intermediate value property. Let $f'(c) = A \neq B = \lim_{x \to c}f'(x)$. Let's take the case when $A < B$. By choosing $\epsilon = (B - A)/2$ and using limit $\lim_{x \to c}f'(x) = B$ we can ensure that there is an $h > 0$ such that $$f'(x) > B - \epsilon = A + \epsilon\tag{1}$$ for all $x \in (c - h, c + h)$. Thus in the interval $(c - h, c + h)$ there is a value of $f'$ less than $A + \epsilon$ namely $f'(c)$ and also as noted above there are values of $f'$ greater than $A + \epsilon$. Hence by intermediate value property $f'$ must take the value $A + \epsilon$ in this interval. But this is not possible because of equation $(1)$ and hence we reach a contradiction. If $A > B$ then we can just interchange the roles of $A, B$ in above argument. Hence we must have $A = B$ and then $f'$ is continuous at $c$.
Note: I have used letter $f$ for the function instead of $h$ as given in question. When dealing with just one function $f$ appears to be a natural choice and I keep the answer that way. I wonder why the question uses $h$ unless it is a part of a larger question and $f, g$ have been used up in other parts of the question.
To expand on user49640's comment, by the MVT $$\frac{h(c + \delta) - h(c)}{\delta} = h'(\xi(\delta)).$$Since $c - |\delta| \le\xi(\delta) \le c + |\delta|$ we have that $\xi(\delta) \to c$ as $\delta \to 0$. Let $L = \lim_{x \to c}h'(x)$. Then $$h'(c) = \lim_{\delta \to 0}\frac{h(c + \delta) - h(c)}{\delta} = \lim_{\delta \to 0}h'(\xi(\delta)) = L.$$
