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Is the following implication true?

Let $f:\mathbb{R} \to \mathbb{R}$ be a continous function, differentiable in all $\mathbb{R}$, besides at most one point $x_0$.

$$f^{(n)}(x) \,\,\mathrm{is \,\,\, continous \,\,\, in \,\,\,} x_0 \implies f \,\,\,\mathrm{is \,\,\, differentiable \,\,\, in }\,\,\, x_0 \,\,\, \mathrm{and} \,\,\, f^{n}(x_0)=\lim_{x \to x_0}f^{n}(x)$$

Where $f^{(n)}$ denotes the n-th derivative of $f$.

I know that such theorem is valid for the first derivative $f'$, but is it valid (as stated) in the case of the n-th derivative?

Paul Frost
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Gianolepo
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    Saying $n$-th derivative is continuous at a point is equivalent to saying that there is a small neighbourhood of $x_0$ in which $f$ is $n$-th differentiable, and $f^{(n)}$ considered as a function is continuous at $x_0$, so the implication must be true – user160738 Dec 06 '16 at 17:32
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    If you know the result for the first derivative consider $f'$ instead of $f.$ Don't you have the result for the second derivative? – mfl Dec 06 '16 at 17:33
  • By even writing $f^{(n)}$ you are already assuming $f$ is $n$ times differentiable at whatever point.... – K.defaoite Sep 23 '21 at 19:15

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The statement of the theorem doesn't make much sense. Take $n=1$ for example. You are assuming $f'$ is continuous at $x_0.$ Well then it's clear that $f'(x_0)$ exists. And by the definition of continuity, $\lim_{x\to x_0} f'(x)=f'(x_0).$ There is nothing going on there, and it is certainly not the result you meant.

Here is what I think the theorem should say: Assume

i) $f:\mathbb R\to \mathbb R$ is continuous everywhere.

ii) $f$ is differentiable on $\mathbb R\setminus \{x_0\}.$

iii) $\lim_{x\to x_0}f'(x)=L$ for some $L\in \mathbb R.$

Then $f'(x_0)=L,$ and $f'$ is continuous at $x_0.$

zhw.
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  • yes @zhw actually I posted this question myself here https://math.stackexchange.com/questions/4256139/nth-time-differentiability-at-a-point-if-limit-of-nth-derivative-exists-at-that and I searched in this site and found similar one – emil agazade Sep 23 '21 at 18:57
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    I'm not sure why you misstate the problem here, when at the other link you stated the problem very well. I have added an answer at the other link. The key idea is to use the MVT, allowing us to get the result without integration. – zhw. Sep 23 '21 at 20:19
  • Sir, it is not my question here and as my reputation is low I can't edit, I just bounted this question to draw attention – emil agazade Sep 24 '21 at 05:51