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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.


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

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  • $\begingroup$ 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

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!

  • $\begingroup$ 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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