Proving differentiability at a point if limit of derivative exists at that point

Suppose that $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous, is known to be differentiable everywhere but $a \in \mathbb{R}$, but that $\lim_{x \rightarrow a}f'(x)=L$. Is it permissible to apply the mean value theorem as follows to show that $f$ is differentiable at $a$?

$\frac{f(a+h)-f(a)}{h}=f'(\theta)$ for some $\theta \in (a, h+a) $ implies that $ \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} = \lim_{\theta \rightarrow a}f'(\theta) = L$ since $\theta$ is sandwiched in between $a$ and $a+h$.

That explaind why the statement is true, but it is not really a proof: since $\theta$ depends upon $h$ and $a$, asserting that $\lim_{\theta\to a}f'(\theta)=L$ makes no sense here.

You can do it as follows. Take $\varepsilon>0$. Now, take $\delta>0$ such that$$0<\lvert x-a\rvert<\delta\implies\bigl\lvert f'(x)-L\bigr\rvert<\varepsilon.\tag1$$Then, if $0<\lvert h\rvert<\delta$, we have $\frac{f(a+h)-f(a)}h=f'(\theta)$ for some $ \theta $ between $a$ and $a+h$. Therefore, $\lvert\theta-a\rvert<\delta$ and then the inequality$$\left\lvert\frac{f(a+h)-f(a)}h-L\right\rvert<\varepsilon$$is a consequence of $(1)$.