1
$\begingroup$

$f\colon (a,b) \to \mathbb{R}$ 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.

  • 0
    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?2010-11-23

2 Answers 2

2

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.

0

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?

  • 0
    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)2010-11-23