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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.

  • 6
    The Mean Value Theorem is your friend.2012-12-13
  • 0
    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.2012-12-14
  • 1
    This is overkill and probably circular logic, but here it comes: This follows imediatelly from L'Hopital.2015-08-19
  • 0
    Possibly somewhat 'useful'/'insightful', (though trivial) remark: The (true) statement in the question implies that if the derivative of a continuous function is discontinuous at a point $a$, then the limit of the derivative approaching $a$ does not exist2017-05-24

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