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This question already has an answer here:

Let $f:[a,b]\rightarrow\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Show that if $\lim_{x\rightarrow a}f'(x)=A$, then $f'(a)$ exists and equals $A$. Use the definition of $f'(a)$ and the Mean Value Theorem.

Definition of a derivative:

Let $I\subset\mathbb{R},f:I\rightarrow\mathbb{R}$, and $c\in I$. We say $L$ is the derivative of $f$ at $c$ if $\forall\varepsilon>0,\exists\delta>0$ such that if $x\in I$ and $0<|x-c|<\delta$, then $|\frac{f(x)-f(x)}{x-c}-L|<\varepsilon$. We say $f$ is differentiable at $c$ and denote it as $f'(c)=L$. We can say $f'(c)=\lim_{x\rightarrow c} \frac{f(x)-f(x)}{x-c} = L$

Mean Value Theorem:

Suppose $f$ is continuous on a closed intervale $I:=[a,b]$ and that $f$ has a derivative in the open interval $(a,b)$. Then there exists at least one point $c\in(a,b)$ such that $f(b)-f(a)=f'(c)(b-a)$

Quite honestly I do not see how I could make these two things work together. Also, if we simply replace $f'(x)$ with its limit definition we would have a limit inside of a limit, and I am not sure how to deal with that either.

Any help in pointing me in the right direction would be greatly appreciated.

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marked as duplicate by Hans Lundmark, user370967, Trevor Gunn, dantopa, Yujie Zha Jun 29 '17 at 15:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This is helpful as I didn't originally realize I could apply the Mean Value Theorem to $(a,x)$ and not just $(a,b)$. However, I do not see the connection to the Squeeze Theorem. This basically says if $f(x)\leq g(x)\leq h(x)$ and $\lim_{x\rightarrow c} f(x) = L = \lim_{x\rightarrow c} h(x)$ then $\lim_{x\rightarrow c} g(x) = L$. Is there an alternate form/some sort of derivation of this I can use to go further? $\endgroup$ – flubsy Dec 10 '15 at 6:24
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We are given that $$\lim_{x\to a}f'(x)=A$$

and know that the MVT guarantees that there exists a number $c\in (a,x)$ such that $$f'(c)=\frac{f(x)-f(a)}{x-a}$$

Note that by the Squeeze Theorem, $x\to a^+ \implies c\to a^+$.

Then, we have

$$\lim_{x\to a^+}f'(c)=\lim_{c\to a^+}f'(c)=f'(a)$$

where we interpret $f'(a)$ as the right-sided derivative. And we are done!

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  • $\begingroup$ What version of the Squeeze Theorem are you using? The one I know says that if $f(x)≤g(x)≤h(x)$ and $\lim_{x→c}f(x)=L=\lim_{x→c}h(x)$ then $\lim_{x→c}g(x)=L$. Is there a different version you are using, or are you adapting this somehow? $\endgroup$ – flubsy Dec 10 '15 at 6:38
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    $\begingroup$ $a<c<x$ So, $\lim_{x\to a}a=a$, $\lim_{x\to a}x=a$. Therefore, $\lim_{x\to a}c=a$. $\endgroup$ – Mark Viola Dec 10 '15 at 14:08
  • $\begingroup$ How do you get the first equality in the last equation? It is true if $f'$ is right-side continuous at $a$. $\endgroup$ – sas Nov 24 '16 at 15:38
  • $\begingroup$ @sas Good question. We know that $f'(c)=\frac{f(x)-f(a)}{x-a}$ for some $c\in (a,x)$. We also are given that $\lim_{c\to a^+}f'(c)=A$ exists. So, the limit quotient does too and is equal to $A$. And we're done! Does that help? Happy Holidays! -Mark $\endgroup$ – Mark Viola Nov 24 '16 at 19:09

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