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I have a question regarding this limit (of a sequence): $$\lim_{n\to \infty} n|\sin(n)|$$ Why isn't it infinite? The way I thought this problem is like this-$|\sin(n)|$ is always positive, and n tends to infinity, so shouldn't the whole limit go to infinity? What is the right way to solve this and why is my idea wrong?

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  • $\begingroup$ Are you asking this as a limit of sequences or as a limit of a real function $f(x)=x| \sinx |$? $\endgroup$ – Marko Karbevski Feb 25 '17 at 18:56
  • $\begingroup$ By the way, as a counter-example to your reasoning: $(1/n^2)$ is always positive as a sequence. Yet $n/n^2 \to 0$. $\endgroup$ – Marko Karbevski Feb 25 '17 at 19:00
  • $\begingroup$ I understand now why the way I thought is wrong, thank you! I don't think it's meant to be a function. How do I calculate this limit then? $\endgroup$ – Jay_Peters Feb 25 '17 at 19:08
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    $\begingroup$ I do not understand why this question was marked as a duplicate. It is not asking for $\liminf_{n\to +\infty} n\left|\sin n\right|$ (whose value is unknown) but just for $\lim_{n\to +\infty} n\left|\sin n\right|$ (that simply does not exist). $\endgroup$ – Jack D'Aurizio Feb 25 '17 at 22:04
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The limit of the sequence $\{n\left|\sin n\right|\}_{n\geq 0}$ as $n\to +\infty$ does not exist. Obviously $\left|\sin n\right|$ is arbitrarily close to $1$ for infinite natural numbers, making the $\limsup=+\infty$. On the other hand, if $\frac{p_m}{q_m}$ is a convergent of the continued fraction of $\pi$ we have $$ \left|p_m -\pi q_m\right|\leq \frac{1}{q_m} $$ and since $\sin(x)$ is a Lipschitz continuous function, the $\liminf$ is finite, by considering $n=p_m$.

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    $\begingroup$ The actual value of the $\liminf$ depends on the elements of the continued fraction of $\pi$ being bounded or not, that is something we do not know at the moment. If the elements are unbounded the $\liminf$ is zero. In any case, it is finite. $\endgroup$ – Jack D'Aurizio Feb 25 '17 at 19:26

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