Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim \limits_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and $$f'(a) = \lim \limits_{x \to a} f'(x).$$
Proof
By definition, $$f'(a) = \lim \limits_{x \to a} \frac{f(a+h) - f(a)}{h}.$$
For sufficiently small $h > 0$ the function $f$ will be continuous on $[a, a + h]$ and differentiable on $(a, a + h)$ (a similar assertion holds for sufficiently small $h < 0$). By the mean value theorem there is a number $\alpha_h$ in $(a, a + h)$ such that $$\frac{f(a + h) -f(a)}{h} = f'(\alpha_h).$$
Now $\alpha_h$ approaches $a$ as $h$ approaches $0$, because $\alpha_h$ is in $(a, a + h)$; since $\lim \limits_{x \to a} f'(x)$ exists, it follows that $$f'(a) = \lim \limits_{h \to 0} \frac{f(a + h) - f(a)}{h} = \lim \limits_{h \to 0} f'(\alpha_h) = \lim \limits_{x \to a} f'(x).$$
The suggestion is to apply a rigorous argument to the last step. To show that $\lim \limits_{h \to 0} \alpha_h = a$ it suffices to choose $\delta = \epsilon$ as then if $|h| < \delta$, then $|\alpha_h - a| < \delta = \epsilon$.
If $\lim \limits_{x \to a} f'(x) = l$, then for $\epsilon > 0$ there is a $\delta > 0$ such that if $|x-a| < \delta$, then $|f'(x) - l| < \epsilon$. Thus if $|h| < \delta$, then $|\alpha_h - a| \leq | (a + h) - a| < \delta$ so $|f'(\alpha_h) - l| < \epsilon$. Thus $\lim \limits_{h \to 0} f'(\alpha_h) = l$.
Is this correct?
