I'm working on the following question:
Assume $f$ is continuous on an interval containing zero and differentiable for all $x \neq 0$. If $\lim_{x\rightarrow 0} f'(x) = L$, show $f'(0)$ exists and equals $L$.
My strategy:
I use the continuity of $f$ to produce an open neighborhood around $0$. In this neighborhood I can find a symmetric closed interval around zero - say $[-\frac{h}{2},\frac{h}{2}]$. By mean value theorem I get $f'(c_1)$ on the interval $(-\frac{h}{2},\frac{h}{2})$. I can use the above fact to create a closed interval containing $c_1$ and $0$ - it will be strictly contained in $(-\frac{h}{2},\frac{h}{2})$. Iterating this procedure will yield a sequence of derivatives $(f'(c_i))$. I'm given that this sequence's limit is $L$. Now I'm stuck.
How do I pass to the limit to show $f'(0)$ takes the value $L$? I don't have any information about the continuity of the derivative.
I know my open interval contains a closed interval that's smaller. What fact am I using here though?