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Let f be a function differentiable on $(a,b)$ and continuous on $c\in(a,b)$. If $c+h \in (a,b)$ then by the mean value theorem $\frac{f(c+h)-f(c)}{h}=f'(c+\theta h)$ for $\theta \in [0,1]$. Let $h \xrightarrow{}0$, then $f'(c+\theta h) \xrightarrow{} f'(c)$ by the above.

My reasoning is the following: $\theta$ is a function of $h$, so in fact it is not true that that $\theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.

user3184807
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    Well under the conditions of your question, it is guaranteed (via mean value theorem) that there is a sequence $c_n$ of values tending to $c$ such that $f'(c_n) \to f'(c) $. What is not guaranteed is that the same result holds for any/every sequence $c_n\to c$ and thus continuity of $f'$ at $c$ can't be inferred. – Paramanand Singh Mar 06 '19 at 05:02
  • That is essentially what my reasoning said – user3184807 Mar 06 '19 at 19:07

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It is not true that $\theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $\theta$'s that work for a given $h.$

We only know $f'(y)\to f'(c)$ as $y\to c$ within the set of $y=x+\theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2\sin(1/x)$ shows.

zhw.
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  • So we can make $\theta$ a function by choosing a value of $h$ and then say the union of all the sequences $\theta(h_n)h_n$ where $h_n \xrightarrow{}0$ over all functions $\theta$ need not be all sequences tending to 0? – user3184807 Mar 04 '19 at 18:27
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    That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2\sin(1/x)$ the sequence $1/(2n\pi)\to 0,$ but this sequence does not arise as $\theta(h_n)h_n$ in the MVT process. – zhw. Mar 04 '19 at 18:35
  • What do you mean by "But the set of such $y$ may not equal any deleted neighborhood of $c$"? Can you clarify it a bit please? – RFZ Apr 13 '21 at 23:14
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If you look at $f(x)=x^2 \sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2x\sin(1/x)-\cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h \in (0,1)$ there exists $\theta(h) \in (0,1)$ with

$$h\sin(1/h)=2h\theta(h)\sin(1/(h\theta(h)))-\cos(1/(h\theta(h))).$$

Notice that once $h$ is small enough, $h\theta(h)$ is forced to stay relatively close to the zeros of $\cos(1/x)$, since

$$|\cos(1/(h\theta(h)))|=|h\sin(1/h)-2h\theta(h)\sin(1/(h\theta(h)))|<3h.$$

This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $\cos(1/x)$. If you instead look at something like $f' \left ( \frac{1}{\pi n} \right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.

Ian
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You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+\theta h) \to f'(c)$ as $h\to0$.

Julián Aguirre
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    No, because OP defined $\theta(h)$ through the MVT in the first place. The point is that you know that $\lim_{h \to 0} f'(c+h\theta(h))=f'(c)$, but why doesn't this imply $\lim_{h \to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $\theta(h)$. – Ian Mar 04 '19 at 17:54
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    That has nothing to do with muy answer. – Julián Aguirre Mar 04 '19 at 17:56
  • Let me spell it out, then. By definition $\lim_{h \to 0} \frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $\frac{f(c+h)-f(c)}{h}=f'(c+h\theta(h))$ for some $\theta(h) \in (0,1)$. That's all OP assumed. Then they send $h \to 0$ in the MVT relation and find $\lim_{h \to 0} f'(c+h \theta(h))=f'(c)$. Now they're wondering why they can't conclude $\lim_{h \to 0} f'(c+h)=f'(c)$. – Ian Mar 04 '19 at 17:57
  • @Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: https://math.stackexchange.com/questions/257907/prove-that-fa-lim-x-rightarrow-afx). – Hans Lundmark Mar 04 '19 at 18:04
  • @HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2\sin(1/x)$, is that $h\theta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT. – Ian Mar 04 '19 at 18:08
  • @Ian: If you can deduce that all information based only on the text in the question, your mindreading abilities are way better than mine... Anyway, if you take a sequence $h_n \to 0$, and a corresponding $\theta(h_n) \in [0,1]$ for each $n$, then of course $\theta(h_n) , h_n \to 0$ too, so that's not where the problem lies. – Hans Lundmark Mar 04 '19 at 18:14
  • The problem is that $\theta(h_n) h_n$ isn't sufficiently arbitrary, it is trapped near the zeros of $\cos(1/x)$. – Ian Mar 04 '19 at 18:16
  • @HansLundmark but the sequence definition of limits require for all $h_n \xrightarrow{}0$ and these may not be covered by sequences $\theta(h_n)h_n$ – user3184807 Mar 04 '19 at 18:16
  • @user3184807: I see what you mean now. I misinterpreted your phrasing in the question and thought that you believed that $\theta(h_n) h_n$ doesn't tend to zero... – Hans Lundmark Mar 04 '19 at 21:33