I am unsure of this proof, and I would appreciate any input about whether my reasoning is on the right track. Due tomorrow btw.

The question is:

Suppose that $\displaystyle f$ is continuous everywhere, $\displaystyle f'$ exists for $\displaystyle x\neq0$, and $\displaystyle \mathop {\lim }\limits_{x \to 0} f'\left( x \right)$ exists. Prove that $\displaystyle f'\left( 0 \right)$ exists.

Professor said to use L'Hôpital's rule if we get stuck.

I started by trying to let $\displaystyle g\left( x \right) = f'\left( x \right)$ and working through an argument using L'Hôpital's rule, but I made some extra assumptions and such which dont seem to be reasonable. Any nudges in the right direction would be great. My work is pretty useless...and ultimately it seems like I wouldnt be proving the right thing anyway. I feel like I am overthinking (as usual) what would be an otherwise simple and elegant proof. If anyone really wants to see my attempt, I will post it, but it is almost certainly incorrect. Thanks in advance.

edit: I know intuitively that it must exist, since $\displaystyle f$ is continuous and $\displaystyle f'$ exists everywhere but $\displaystyle 0$ (for sure). And since the two-sided limit exists at $\displaystyle 0$, either $\displaystyle f'$ is continuous at $\displaystyle 0$ or has a removable discontinuity there, but it must be defined there because $\displaystyle f$ is continuous at $\displaystyle 0$ and does not have a corner there. There cannot be a corner at $\displaystyle f \left( 0 \right)$ because the two-sided limit exists, so we know that the derivative must exist there (as $\displaystyle f$ is continuous without a corner)... but I do not see how L'Hôpital would come into play in a formal proof?