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Let $f:[0,\infty)\to\mathbb{R}$ be a function. Suppose that $f$ is bounded, differentiable and satisfies the following inequality: $$f(x)f'(x)\geq \sin(x)\text{ for all }x\in[0,\infty).$$ Does there exist $\lim\limits_{x\to\infty}f(x)$?

I suppose that the statement is false since I don't see how the inequality will help us prove the limit existence. However, I have not found out any counterexample yet.

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1 Answer 1

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Take $f(x) = \sqrt{3-2\cos(x)}$. Then $f$ is bounded, differentiable and $$ f(x)\,f'(x) = \frac{1}{2} (f(x)^2)' = (\tfrac{3}{2}-\cos(x))' = \sin(x) $$ but $f$ has no limit at $+\infty$.

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    $\begingroup$ Thanks!! Would you tell me how you found this function? $\endgroup$
    – Lord Vader
    Sep 14, 2021 at 17:01
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    $\begingroup$ Yes. The first step is to replace $f\,f'$ by $g' = (f^2/2)'$. Then solve $g' = \sin$ with the constraint that $g$ is strictly positive ;) Of course you can replace $3$ by any $C > 2$. $\endgroup$
    – LL 3.14
    Sep 14, 2021 at 17:04
  • $\begingroup$ I get it. Thanks!! $\endgroup$
    – Lord Vader
    Sep 14, 2021 at 17:09
  • $\begingroup$ Great example... $\endgroup$
    – Randall
    Sep 14, 2021 at 17:47

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