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Suppose $f: (0, 1) \rightarrow {\rm I\!R}$ is continuous, with $c \in (0, 1)$, and suppose that $f'$ exists $\forall x \in (0, 1)$\{$c$}. Prove that if $\lim_{x\to\ c} \ f'(x) = K$ then $f'(c) = K$.

$$\lim_{x\to\ c} \ f'(x) = K \Rightarrow \forall \epsilon > 0, \exists \delta > 0 \text{ such that } |f'(x) - K| < \epsilon \text{ for } x \in (c-\delta, c+\delta)$$

Since $f$ is continuous and differentiable on the open interval we can use the Mean Value Theorem on $[x, c]$ where $x \in (c-\delta, c)$.

So $\exists x_1 \in (x, c)$ such that $f'(x_1) = \frac{f(x) - f(c)}{x-c}$.

By our limit definition $|f'(x_1) - K| = |\frac{f(x) - f(c)}{x-c} - K| < \epsilon$.

$$\Rightarrow \left|\frac{f(x)-f(c)}{x-c} - K\right| < \epsilon \text{ for } x \in (c-\delta, c)$$

We can use the MVT again on $[c, x]$ where $x \in (c, c+\delta)$, find an $x_2 \in (c, c+\delta)$ and get similar result.

$$\Rightarrow \left|\frac{f(x)-f(c)}{x-c} - K\right| < \epsilon \text{ for } x \in (c-\delta, c+\delta)$$

$$\Rightarrow \lim_{x\to\ c} \frac{f(x)-f(c)}{x-c} = K$$

$$\Rightarrow f'(c) = K$$

I'm not sure that I did this right - I'm not too comfortable with proofs yet (especially unfamiliar with using the MVT in proofs). I didn't use the fact that the domain of f is (0,1) which is concerning.

Michael Hardy
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user20354139
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1 Answers1

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You have the right approach.

You only need to use the MVT once:

Suppose you have chosen $\epsilon, \delta $ as above. If we take $|h| < \delta$ then for some $\xi \in [c,c+h]$ we have $|{f(c+h) -f(c) \over h }-K| = |f'(\xi)-K| < \epsilon$.

copper.hat
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    I think his concern was that one needs a generalized version of the interval if $h<0$. He probably doesn't have that. (Did not downvote btw.) – Lukas Betz Feb 04 '17 at 23:20
  • @LeBtz: I glossed over that entirely, I believe you are correct. – copper.hat Feb 04 '17 at 23:21
  • @LeBtz: Thanks, I know who the downvoter is (at least the userid). This one is justifiable, at least. – copper.hat Feb 04 '17 at 23:22
  • Yeah I used it twice because $[c, c + h]$ doesn't fit our convention for writing intervals if $h < 0$. – user20354139 Feb 04 '17 at 23:38
  • @antsatsui: Your answer makes perfect sense then. I use the notation $[x,y]$ to mean all points on and in between $x,y$, so it has no sign, as such. – copper.hat Feb 04 '17 at 23:40
  • Is there any reason that problem specified $f$ on $(0,1)$? It's odd that they would give unnecessary information. – user20354139 Feb 04 '17 at 23:41
  • @antsatsui: I can't think of a reason since they specify $c \in (0,1)$ anyway. Perhaps to ensure that an unambiguous two-sided limit exists? – copper.hat Feb 04 '17 at 23:42