# prove that $f'(a)=\lim_{x\rightarrow a}f'(x)$.

Let $f$ be a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$.
Suppose that $\lim_{x\rightarrow a}f'(x)$ exists.
Then, prove that $f$ is differentiable at $a$ and $f'(a)=\lim_{x\rightarrow a}f'(x)$.

It seems like an easy example, but a little bit tricky.
I'm not sure which theorems should be used in here.

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Using @David Mitra's advice and @Pete L. Clark's notes
I tried to solve this proof. I want to know my proof is correct or not.

By MVT, for $h>0$ and $c_h \in (a,a+h)$ $$\frac{f(a+h)-f(a)}{h}=f'(c_h)$$
and $\lim_{h \rightarrow 0^+}c_h=a$.

Then $$\lim_{h \rightarrow 0^+}\frac{f(a+h)-f(a)}{h}=\lim_{h \rightarrow 0^+}f'(c_h)=\lim_{h \rightarrow 0^+}f'(a)$$

But that's enough? I think I should show something more, but don't know what it is.

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The Mean Value Theorem is your friend. –  David Mitra Dec 13 '12 at 14:10
Your phrasing is off. At the start, you want to say "Let $h>$. By the MVT, there is a $c_h\in(a,a+h)$... . Then say "Now note that $\lim_{h\rightarrow 0^+} c_h=a$. The only place where you might need to provide additional justification is the last equality in the last displayed equation. But this should be easy for you. –  David Mitra Dec 14 '12 at 15:47

Some hints:

Using the definition of derivative, you need to show that $$\lim_{h\rightarrow 0^+} {f(a+h)-f(a)\over h }$$ exists and is equal to $\lim\limits_{x\rightarrow a^+} f'(x)$.

Note that for $h>0$ the Mean Value Theorem provides a point $c_h$ with $a<c_h<a+h$ such that $${f(a+h)-f(a)\over h } =f'(c_h).$$

Finally, note that $c_h\rightarrow a^+$ as $h\rightarrow0^+$.

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Why not using l'Hôpital's theorem? It applies in this case. Of course its proof uses Cauchy's theorem, which is equivalent to the MVT. –  egreg Apr 29 '13 at 12:06