I want to calculate the limit: lim supn→∞sin(n)\displaystyle\limsup_{n\to\infty} \sin(n)

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




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



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


1 Answer


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