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Suppose that $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Then, by the mean value theorem, $\forall x\in(a,b]$ $\exists c\in(a,x)$ such that $f '(c)=(f(x)-f(a))/(x-a)$.

It seems to me that, since $c\in(a,x)$, as $x$ goes to $a$ from the right, $c$ must also "go to" $x$ and thus $f'(c)$ goes to $f'(x)$. Is this correct? If not, where have I gone wrong?

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The point c depends on x, and is different for each value of x. – Calvin Lin Jan 5 '13 at 2:42
Yes, $c$ depends on $x$ and can vary as $x$ varies. Indeed, this fact is what I was using to try to "trap" $c$. As $x$ gets very close to $a$, $c$ must also be getting very close to $x$. This is what I was relying on (rightly or wrongly) in my argument. – jim Jan 5 '13 at 3:03

Well, I think you're right if we assume say that the derivative is continuous. Applying L'Hospital we'd get

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

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Ok, thanks. However, we don't know that $f'(a)$ exists. – jim Jan 5 '13 at 2:57
@Don I would agree with your equation if there was no $f'(a)$ in there. In fact, that is all we can say, since $f'(a)$ might not exist, if the limit tends to infinity, like in $f(x) = \sqrt{x}$. – Calvin Lin Jan 5 '13 at 3:14
Indeed so, @CalvinLin...something's still missing here, and I think that as the question's given we can't reach a final conclusion: if $\,f'(a),$ exists and $\,f'(x)\,$ is continuous we can do as shown above, yet $\,f(x)=\sqrt x\,\,,\,\,a=0\,$ is a nice example that the above can miserably fail, as $\,(\sqrt x)'\,$ doesn't exists at $\,a=0\,$ – DonAntonio Jan 5 '13 at 10:55

Two issues here. First, what does it mean that $c$ "goes to" $x$? A limit is a number, not a variable. And $x$ is a variable. It would be correct to say that $c$ goes to $a$.

If $f'(a)$ exists (which you did not assume), then using your argument we can conclude the following: there are numbers $c$ arbitrarily close to $a$ for which $f'(c)$ is arbitrarily close to $f'(a)$. This can be put more precisely in the language of sequences: there exists a sequence $c_n\to a$ such that $f'(c_n)\to f'(a)$

However, even then it does not follow that $\lim_{c\to a} f'(c)=f'(a)$. The definition of limit requires us to investigate all values of $c$ that are close to $a$. We looked only at those $c$ which the Mean Value Theorem threw at us.

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