I am not entirely sure how to go about this one.

lim supn→∞sin(n)\displaystyle\limsup_{n\to\infty} \sin(n)

I am assuming \sup\sup means supremum, in which case since \sin\sin can only be in range of <-1,1><-1,1>, supremum is 11, which makes the limit 11. Is this correct?

Edit: I misunderstood the task, corrected it, but that makes the above comment irrelevant

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The expression is constant, so the value is \sin n\sin n. Did you mean to have xx as the limit variable?

– copper.hat

Oct 20 at 18:04

The answer is 11. \sin n\sin n is dense in (-1,1)(-1,1).

– MathematicsStudent1122

Oct 20 at 18:05

Do you mean \displaystyle\limsup_{n\to\infty} \sin(n)\displaystyle\limsup_{n\to\infty} \sin(n)? (wiki)

– arkeet

Oct 20 at 18:06

As written, \sin(n)\sin(n) is independent of the limit variable, so it’s constant – the answer would be \sin(n)\sin(n).

– arkeet

Oct 20 at 18:07

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Possible duplicate of Showing \sup \{ \sin n \mid n\in \mathbb N \} =1\sup \{ \sin n \mid n\in \mathbb N \} =1

– Byron Schmuland

Oct 20 at 18:15

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1 Answer

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The sequence \{e^{in}\}_{n\geq 0}\{e^{in}\}_{n\geq 0} is dense in the unit circle, and the function z\to \text{Im}(z)z\to \text{Im}(z) from the unit circle to the [-1,1][-1,1] interval is a continuous function, preserving density. It follows that 11 is a limit point for the sequence \{\sin n\}_{n\geq 0}\{\sin n\}_{n\geq 0}, so your \limsup\limsup equals \color{red}{\large 1}\color{red}{\large 1}.

Notice that the range of the sine function is [-1,1][-1,1] is not enough to reach that conclusion.

For instance, the range of the functions f(t)=\frac{2t}{1+t^2}f(t)=\frac{2t}{1+t^2} or g(t)=\sin(\pi t)g(t)=\sin(\pi t) is also [-1,1][-1,1], but

\limsup_{n\to +\infty}\frac{2n}{1+n^2}=\limsup_{n\to +\infty}\sin(\pi n)=\color{red}{0}. \limsup_{n\to +\infty}\frac{2n}{1+n^2}=\limsup_{n\to +\infty}\sin(\pi n)=\color{red}{0}.

Equivalently, n-2 \pi \lceil {n \over 2 \pi} \rceiln-2 \pi \lceil {n \over 2 \pi} \rceil is dense in [0,2 \pi][0,2 \pi].

– copper.hat

Oct 20 at 18:19

@copper.hat: sure, that is also underlined in the linked proof.

– Jack D’Aurizio

Oct 20 at 18:20

That proof looks rather complicated for this case. Wouldn’t it be enough to show that \sup \sin(n)\sup \sin(n) is always 11, hence the limit is 11?

– Mykybo

Oct 20 at 18:28

@Mykybo: but \sup_{n}\sin(n)=1\sup_{n}\sin(n)=1 still follows from density. What alternative (simpler) approach do you have in your mind?

– Jack D’Aurizio

Oct 20 at 18:33