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

=================

2

 

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

1

 

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
1

=================

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