for any real number t between -1 and 1, does there exist a infinite sequence of positive integers $x_{n}$ such that $\lim_{n\to\infty}{\sin(x _{n})}=t$ ?
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$\left{\sin{n} \mid n \in\mathbb{N}\right}$ is dense in $[-1,1]$. You will find a proof here. – rtybase Sep 26 '20 at 15:14
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Yes. The set of points $\{e^{in}\}$ is dense on the unit circle in the complex plane, so you can find a subsequence converging to any point there.
Ethan Bolker
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Why is the set of points ${e^{in}}$ dense on the unit circle in the complex plane? – 김재현 Sep 26 '20 at 15:10
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That follows from the equidistribution theorem in the wikipedia link. – Ethan Bolker Sep 26 '20 at 15:11
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Hints: as ,the set of subsequential limits of $\{\sin(n)\}_{n}$ is dense in $[-1,1]$ , so, for every $t\in [-1,1]$ , there is a subsequence $\{x_n\}$ of the sequence $\{n\}$ such that $\sin(x_n)=t$
A learner
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