Is it possible to prove the following without using the mean value theorem:
If $f$ is differentiable on an interval containing $0$ and if $\lim_{x \to 0} f'(x) = L$ then $f'(0) = L$.
I have tried (see here) but I'm running out of ideas and I'm wondering if it's possible at all.
The MVT is really the simplest tool you can use. Here an "alternative" proof.
By considering $f(x)-f(0)-xL$ we reduce to the case where $f(0)=0$ and $f'(0)=0$.
Claim: there is a sequence of points $x_n\to0$ s.t. $f'(x_n)\to 0$. Since $f'(x)$ has a limit when $x\to 0$ the claim concludes.
Proof of the claim. Suppose the contrary. Therefore in a neighborhood of $0$ we have no local maxima or minima, because there the derivative is zero.
REMARK: This is equivalent to reproving the MVT!!!
Thus $f$ is monotone near zero, say increasing. Again by contradicting the claim, we get $f'(x)>\epsilon$ near zero. Thus $f(x)>\epsilon x$ near zero (this is again another hidden instance of the MVT). This makes impossible to have $f'(0)=0$.
But as David Mitra said, the MVT is a simple and useful result, why don't use it directly?