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I ran across this problem in my Analysis class and can't seem to come up with a good solution. Here's the question: Show that if $\lim_{x\to a}f'(x) = A$ then $f'(a)$ exists and equals $A$. $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$.

I tried applying the Mean Value theorem like this: There exists a $c \in (a, x)$ st $f(x)-f(a)=f'(c)(x-a) \implies \frac{f(x)-f(a)}{x-a}=f'(c) \implies f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{c\to a}f'(c)=A$

However, I think this is wrong. I am pretty sure the Mean Value theorem is needed, but my application of it seems to be incorrect. Would somebody be able to point me in the right direction?

Keith Z
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1 Answers1

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Applying $\lim \limits_{(x\text{ or }c)\to a}$ as you did doesn't feel right.

Hint: You know that $$\mathop{\forall}_{\varepsilon >0}\mathop{\exists}_{\delta >0}\mathop{\forall}_{c\in ]a,b[}\left(c-a<\delta\implies |f'(c)-A|<\varepsilon\right)\tag 1$$

and you wish to prove that

$$\mathop{\forall}_{\varepsilon >0}\mathop{\exists}_{\delta >0}\mathop{\forall}_{x\in ]a,b[}\left(x-a<\delta\implies \left|\dfrac{f(x)-f(a)}{x-a}-A\right|<\varepsilon\right)\tag 2$$

Take $\varepsilon >0$, let $\delta$ be as in $(1)$ and let $x\in ]a,b[$ be such that $x-a<\delta$. Now use the mean value theorem on $f_{|[a,x]}$.

Git Gud
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  • Okay so I think I got it. By the MVT we have that $f'(c)=\frac{f(x)-f(a)}{x-a}$. Since $x-a<\delta \implies c-a<\delta$ we can substitute f'(c) in (2) and have it be true because of (1). – Keith Z Dec 16 '14 at 18:36
  • @KeithZ Yes, that's correct. – Git Gud Dec 16 '14 at 18:37