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Suppose we have a function $f : \mathbb{R}^+ \to \mathbb{R}$. It seems intuitive to me that if $\lim_{x \to 0} f(x)$ exists, then $\lim_{x \to 0} x f'(x) = 0$. I suspect that for real functions, there may be pathological counterexamples to this, but at least for analytic functions (where $0$ may be on the boundary of the analytic disk) then it should be true.

In the analytic case, I can give an argument for this that would convince a typical physicist like myself. $f(x)$ cannot have an essential singularity at $0$, because $\lim_{x \to 0} f(x)$ would not exist. So as $x \to 0$, it scales like $f(x) = c + O( x^\alpha )$ for some constant $c$ and $\alpha > 0$, since the limit exists. Therefore, $x f'(x) = O( x^\alpha ) \to 0$.

Two questions:

  1. Is this true, and if so, how generally valid is it?

  2. Is there a simpler proof that does not rely on scaling arguments? This seems like the kind of problem that one would typically address with a fairly general theorem like L'Hospital's rule. Another strategy is to write $xf'=(xf)'-f$, but this just shifts the burden of the problem to showing that $\lim_{x \to 0}[xf(x)]'=\lim_{x \to 0} f(x)$. I am guessing that the solution is totally obvious and I am just not seeing it.

sasquires
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    it's false. try $f(x) = x \sin \frac{1}{x}$. $xf'(x) = x \sin \frac{1}{x} - \cos \frac{1}{x}$ – D F Jul 12 '18 at 17:01
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    @DF : Maybe the OP meant, that $f'(0)$ exists. Then it's trivial what he says. Otherwise your example is of course correct. – user90369 Jul 12 '18 at 17:11
  • @DF I think that in your example, $f'(x)=x \sin\frac{1}{x} - \color{red}{ \frac{1}{x} } \cos \frac{1}{x}$. – sasquires Jul 12 '18 at 18:23
  • @sasquires you can use wolframalpha in order to check your calculations... – D F Jul 12 '18 at 18:25
  • @DF Just in case, I did check, and Wolfram Alpha agrees with me. You forgot to use the chain rule on $\frac{1}{x}$. – sasquires Jul 12 '18 at 18:28
  • @sasquires lol. I didn't forget anythinng. $$f'(x) = (x)' \sin \frac{1}{x} + x (\sin \frac{1}{x})' = 1 \times \sin \frac{1}{x} + x \times (\frac{1}{x})' \cos \frac{1}{x} = 1 \times \sin \frac{1}{x} + x \times - \frac{1}{x^2} \cos \frac{1}{x} = \sin \frac{1}{x} - \frac{1}{x} \cos \frac{1}{x}$$ and the next step is $$x \times f'(x) = x \times (\sin \frac{1}{x} - \frac{1}{x} \cos \frac{1}{x}) = x \sin \frac{1}{x} - \cos \frac{1}{x}$$! As I wrote! – D F Jul 12 '18 at 18:34
  • @DF Sorry, my mistake; not sure what I was thinking this morning. – sasquires Jul 12 '18 at 20:02
  • @sasquires that's all right :) – D F Jul 12 '18 at 20:06

1 Answers1

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You already suspect it may not be true in the real case. The misinterpretation of your question has already provided a few counterexamples which can only arise when the limit of $xf'(x)$ does not exist.

One thing to note in the real case is if $f$ is differentiable in an excluded neighborhood of $0$, $\lim_{x \to 0} f(x) = 0$ and $\lim_{x \to 0} x f'(x)$ exists, then it must be $0$. By L'Hopital's rule

$$0 = \lim_{x \to 0+} \frac{f(x)}{\log x} = \lim_{x \to 0+} xf'(x) , \\ 0 = \lim_{x \to 0-} \frac{f(x)}{\log (-x)} = \lim_{x \to 0-} -xf'(x) $$

The form of L'Hopital's rule that applies here is where $g(x) \to \infty$ implies that $\lim_{x \to 0} f(x)/g(x) = \lim_{x \to 0} f'(x)/g'(x)$ (when the limit on the RHS exists) without the assumption that $f(x) \to \infty$. Apparently this is not commonly known. See Case 2 in General Proof here to confirm.

RRL
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  • Great, I thought I should have been able to do it with L'Hospital's rule, but I didn't think of using the logarithm in the denominator, and I did not know about this form of L'Hospital's rule. So this addresses most of the question. But if we take the case of an complex analytic function (where $0$ may be on the boundary of the disk of analyticity), do we gain anything? Is there a problem with my hand-wavy argument? If it is true for analytic functions, is there a simpler argument? – sasquires Jul 12 '18 at 18:18
  • I'm accepting the answer and will post a new question about the analytic case. – sasquires Jul 17 '18 at 21:55
  • Do you have any idea whether this holds in other absolute-valued fields? If $\lim_0f$ and $\lim_0xf'$ exist, must the latter be $0$? This result might be applicable to my Question https://math.stackexchange.com/questions/4690961/in-characteristic-2-can-a-function-have-a-non-zero-second-derivative – mr_e_man May 09 '23 at 01:49