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$f\colon (a,b) \to \mathbb{R}$ is continuous, with finite derivatives everywhere in $(a,b)$, except maybe at $c$.
If $$\lim_{x\to c}f'(x) = B,$$ show that $f'(c)$ exists and equals $B$.

I'm not sure where to start on this. I've tried using the definitions of continuity and f' but it isn't working out

Well I started out with $$\lim_{x\to c}\frac{f(x)-f(c)}{x-c} = f'(c)$$ is equivalent to saying
given $\epsilon\gt 0$ there exists $\delta\gt 0$ such that
$$|x-c| \lt \delta \Longrightarrow \left|\frac{f(x)-f(c)}{x-c} - f'(c)\right| \lt\epsilon.$$ I was trying to prove $f'(c)$ must exist since $f$ is continuous but I feel like I'm assuming what I'm trying to prove.

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What do you mean it isn't working out? Please add your working to the question, we can help you better that way. –  Aryabhata Nov 23 '10 at 18:40
    
Hint: mean value theorem –  Robin Chapman Nov 23 '10 at 18:42
    
Why do we care that f has a finite derivative everywhere except maybe at c? Using MVT, doesn't it only matter that the derivative exists? –  JimJones Nov 23 '10 at 18:50
    
Can I adjust my interval (a,b) to (c-e, c+e)=C with C a subset of (a,b). Then prove that for every c in (a,b) I can find an interval C such that f'(c)=f(c+e)-f(c-e)/(c+e-(c-e)) using MVT? –  JimJones Nov 23 '10 at 18:57

2 Answers 2

up vote 2 down vote accepted

You are right that you are kind of assuming what you want, because you cannot even write down $f'(c)$ until you prove it exists; this can be fixed if you replace $f'(c)$ by a specific limit, and the specific limit you want is probably $B$. But even so ran into a bit of an alley, so let me help you out. Here is the beginning of half a proof, following the Hint of Robin Chapman:

Consider first the limit as $x\to c^+$. For any $h\gt 0$, the function $f(x)$ is continuous on $[c,c+h]$, and is differentiable on $(c,c+h)$, so by the Mean Value Theorem there exists a point $d_h$ (which depends on $h$) such that $$\frac{f(c+h)-f(c)}{h} = f'(d_h).$$ Therefore, we have that \begin{align*} \lim_{x\to c^+}\frac{f(x)-f(c)}{x-c} &= \lim_{h\to 0^+}\frac{f(c+h)-f(c)}{h}\ &=\lim_{h\to 0^+}f'(d_h). \end{align*}

So we've managed to turn the (one-sided) limit that will tell us whether $f$ is differentiable at $c$ into a limit about $f'(x)$. Luckily, we have information about limits of $f'(x)$. So perhaps we can leverage that information into an actual value for this limit? If so, perhaps we can try doing the same thing for the limit from the left (as $x\to c^-$), so that the two together shows us that $$\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$ exists; and will tell us what the value of $f'(c)$ must be in that case.

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So what can go wrong? We can certainly define a function g(x)=something nice except when x=c, something different at c. Then $\lim_{x\rightarrow c} g(x) \neq g(c)$ Can you rule out this behavior in $f'(x)$? Or maybe $f'(x)$ has a step at $c$. Why can't it?

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f'(x) can't have a step at c since lim as x -> c of f'(x) exists. But I don't understand why lim as x -> c of f'(x) must equal f'(c) –  JimJones Nov 23 '10 at 19:19

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