Is the right-hand derivative equal to the right-hand limit of the derivative? Let $f(x)$ be a function on the interval $[a,b]$ which is differentiable on $(a,b)$. Is it true that
$$f'_+(a)=\displaystyle\lim_{x\to a^+}f'(x)$$
if both limits exist? Darboux's theorem seems to imply that it is indeed the case, but my idea of proof is somewhat fishy (uses an ``odd extension" of $f(x)$, etc.) Can anyone confirm or disprove that I'm right? Thanks.
Here $f'_+(a):=\displaystyle\lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}$.
 A: As hinted by @Ricky Demer, the MVT is what you want to use. Here's why. 
Suppose that $f$ is differentiable on $(a,b)$ and that  $\lim_{x\to a+} f'(x)=L\in{\mathbb R}$. Let $(h_n)$ be a positive sequence tending to $0$. Then it follows from the MVT that for each $n$ there exists $c_n \in (a,a+h_n)$ such that
$$\frac{f(a+h_n) - f(a)}{h_n}= f'(c_n).$$
By letting $n\to\infty$, the righthand side converges to $L$, and therefore so does the lefthand side. Since this limit is independnet of the sequence $(h_n)$, it folllows that $f_+'(a)$ exists and coincides with $\lim_{x\to a+} f'(x)$. 
Also observe that this argument shows that the only discontinuities of a derivative are of the second kind (never removable or first kind (limit exists but is different from value)). 
A: Your answer is right when the function is differentible, that is $f'$ is exists.
In this case bouth $f'_+$ and $f'_\_$ are exists and equal and there is no problem.
If $f$ is only have right derivative there is no $f'$ and it is not the case. 
