Can you please check my proof of this limit of derivative I proved the following:

If $f$ is differentiable on an interval containing $0$ and if $\lim_{x
> \to 0} f'(x) = L$ then $f'(0) = L$.

Please can you tell me if my proof is correct:
By contradiction assume $f'(0) \neq L$. Without loss of generality assume $f'(0) > L$. Also without loss of generality assume $L>0$. Since $\lim_{x \to 0} f'(x) = L$ there exists $\delta > 0$ such that 
$$ 0 < |x|<\delta \implies |f'(x) -L|<{L\over 2}$$
which implies that for $ 0 < |x|<\delta$ it holds that $0 \le f'(x)<{L\over 2}$. Fix $x_0>0$ with $f'(x_0)< {L \over 2}$. Then 
$$f'(0) > L > {L\over 2}>f'(x_0) $$
but there is no $c \in (0,x_0)$ with $f'(c) = L$ which is a contradiction to Darboux's theorem. Hence it must holds that $f'(0) = L$.
 A: Hmm, I do have my doubts how you arrive at the conclusion $f^\prime(x) < \frac{L}{2}$. I get $f^\prime(x)< \frac{3L}{2}$ or $f^\prime(x) >\frac{1}{2}$ Also, assuming you can show that, you should at least explain why you may assume that $L\neq 0$, since you make explicit use of that fact (this is of course easy, you can, e.g. just add $ax$ to $f$) 
To be honest, I also think the approach is too complicated. I hope you don't mind if I propose an easier approach: by the mean value theorem, for each $x$ near $0$ there is $c\in (0,x)$ (or $\in (x, 0)$, depending on the sign of $x$), such that
$$\frac{f(x)-f(0)}{x} = f^\prime(c)$$
Now if you let $x\rightarrow 0$ you get the claim immediately from the definition of differentiability and the assumption about the convergence of $f^\prime$ (since, of course, the corresponding $c\rightarrow 0$, as well).
This path of thought actually incited me to write my first comment, now deleted (which said you need to show first that $f^\prime(0)$ exists). With this reasoning you don't need to know that $f$ is differentiable at $0$, you get it for free.
