I'm trying to solve a problem that goes as follows :
Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If $\lim_{x\to{a}} f^{'}(x)$ exists, show that $(1)$, $f$ is differentiable at $a$, and $(2)$, that $f^{'}(a)=\lim_{x\to{a}} f^{'}(x)$
I know that this has been posted on the stack exchange before, notably here and here, but I would like to confirm my understanding of their approach and a question I have a have about the differentiability portion of the problem.
We can firstly see that this problem sets us up to use the MVT, and so we know $\exists$ some $c$, s.t. $f^{'}(c)$ is the secant line between $a$ and $x$, thus we can write this as $f^{'}(c)=\frac{f(x)-f(a)}{x-a}$. We also know that the problem allows us to assume that $\lim_{x\to{a}} f^{'}(x)$ exists, and therefore equals some limit $L$. Now for the proof of $(2)$, (my interpretation of their sol. is that:) we notice that this MVT interval can be shrunk as we send $x\rightarrow a$, therefore we are forcing this point $c$ to $a$ as $x$ is sent to $a$. Explicity, we get
\begin{align} f^{'}(x) = \lim_{x\to{a}} \frac{f(x)-f(a)}{x-a} \implies f^{'}(x) = \lim_{x\to{a}} f^{'}(c) \end{align} thus as $x\rightarrow a$, we get: \begin{align} \lim_{x\to{a}} f^{'}(x) = \lim_{c\to{a}} f^{'}(c) = f^{'}(a) \implies \lim_{x\to{a}} f^{'}(x) = f^{'}(a) \end{align}
My interpretation of this, is that by limiting this interval in which the mean value theorem applies, we're sort of forcing the secant line to be the derivative of the function at $a$.
My problem lies more in part $(1)$, where I need to prove differentiability (side question: is $(1)$ needed for $(2)$?). I'm having trouble approaching this from an $\epsilon-\delta$ perspective. I know that differentiability at $a$ implies the existence of $\lim_{x\to{a}} \frac{f(x)-f(a)}{x-a}$. I don't immediately see anything I can use from before to say that $f$ will be differentiable at $a$, so I attempted an $\epsilon-\delta$ proof as follows: \begin{align} |\frac{f(x)-f(a)}{x-a}-L|=|f^{'}(c)-L| \text{(from MVT)} \end{align} I'm not sure how to proceed from here. Is there something obvious about this that allows me to find $\delta$ immediately? Any help would be greatly appreciated!
