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Let $f(x)=x^{-1}\sin\left(x^3\right)$. It's easy to check that that the limit of $f$ at $+\infty$ is $0$ and that $f'$ doesn't have a limit. However, applying this trick, we can deduce that $\lim_{+\infty}f'(x)=0$. How to resolve this contradiction?

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user5402
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2 Answers2

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That trick requires the assumption that $\lim_{x \to \infty} f(x) + f'(x)$ exists.

Robert Israel
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  • $\lim_{x\to\infty}\ f(x)\ =\ \lim_{x\to\infty}\frac{e^x\ f(x)}{e^x}$ is obvious. $\lim_{x\to\infty}\frac{e^x\ f(x)}{e^x}\ =\ \lim_{x\to\infty}\frac{e^x\ (f(x)+f,'(x))}{e^x}$ is just L'Hopital's rule. $\lim_{x\to\infty}\frac{e^x\ (f(x)+f,'(x))}{e^x}\ =\ \lim_{x\to\infty}, (f(x)+f'(x))$ is also obvious. So in which step we assumed that $\lim_{x \to \infty} f(x) + f'(x)$ exists? – user5402 Nov 15 '15 at 18:18
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    The problem with that is that $\lim_{x\to\infty} \frac{e^xf(x)}{e^x}$ is not an indeterminate $\infty/\infty$ form because $e^xf(x)$ does not tend to $\infty$ (for example, because it is zero infinitely often). – user281392 Nov 15 '15 at 18:28
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    @whatever L'Hopital's rule says that when the limit you get after differentiating (in this case $\lim f(x)+f'(x)$) exists, then the original limit exists and the two are equal. When the second limit doesn't exist, we can't say anything about the first one. – Kitegi Nov 15 '15 at 19:08
  • @Farnight Yes I totally forgot that. – user5402 Nov 15 '15 at 19:11
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If you use a Proposition, Lemma or Theorem, you always need to check if your hypothesis is fullfilled. In this case, your trick says: if $f+f' \to L$ (where $L\in\mathbb{R}$) as $x\to \infty$, then $f\to L$,$f'\to 0$. But in your case, the limit of $f'$ does not exist so the limit of $f+f'$ does not exist either.

For more detailed understanding, you see that you can not follow the same path of argumentation as in the proof with your specific example, as $lim_{x\to\infty}\ e^xf(x)$ does not exist. This means that you can not have the equality $\rm \lim_{x\to\infty}\frac{e^x\ f(x)}{e^x}\ =\ \lim_{x\to\infty}\frac{e^x\ (f(x)+f\,'(x))}{e^x}\ \qquad$ as in the proof because it uses L'Hospital and you do not have the indeterminate form "$\infty/\infty$" for $lim_{x\to\infty}\frac{e^x\ f(x)}{e^x}$ (because the limit of the numerator does not exist).

I hope I could help.

derthomas
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