I want to calculate the limit: $\displaystyle\limsup_{n\to\infty} \sin(n)$

Question

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

$\displaystyle\limsup_{n\to\infty} \sin(n)$

I am assuming $\sup$ means supremum, in which case since $\sin$ can only be in range of $<-1,1>$, supremum is $1$, which makes the limit $1$. 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 $\sin n$. Did you mean to have $x$ as the limit variable? — copper.hat, Oct 20 '16 at 18:04
Do you mean $\displaystyle\limsup_{n\to\infty} \sin(n)$? (wiki) — arkeet, Oct 20 '16 at 18:06
As written, $\sin(n)$ is independent of the limit variable, so it's constant - the answer would be $\sin(n)$. — arkeet, Oct 20 '16 at 18:07
@arkeet you are actually right, now I am completely lost, gotta go study what it means — Mykybo, Oct 20 '16 at 18:08
@Mykybo, how about editing your question to state the problem correctly and formatting appropriately? — user8960, Oct 20 '16 at 18:10
Possible duplicate of Showing $\sup { \sin n \mid n\in \mathbb N } =1$ — , Oct 20 '16 at 18:15

score 2 · Accepted Answer · edited Apr 13 '17 at 12:19

2

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

Notice that the range of the sine function is $[-1,1]$ is not enough to reach that conclusion.
For instance, the range of the functions $f(t)=\frac{2t}{1+t^2}$ or $g(t)=\sin(\pi t)$ is also $[-1,1]$, but $$ \limsup_{n\to +\infty}\frac{2n}{1+n^2}=\limsup_{n\to +\infty}\sin(\pi n)=\color{red}{0}.$$

edited Apr 13 '17 at 12:19

Community

1

answered Oct 20 '16 at 18:16

Jack D'Aurizio

353,855

Equivalently, $n-2 \pi \lceil {n \over 2 \pi} \rceil$ is dense in $[0,2 \pi]$. – copper.hat Oct 20 '16 at 18:19
@copper.hat: sure, that is also underlined in the linked proof. – Jack D'Aurizio Oct 20 '16 at 18:20
That proof looks rather complicated for this case. Wouldn't it be enough to show that $\sup \sin(n)$ is always $1$, hence the limit is $1$? – Mykybo Oct 20 '16 at 18:28
@Mykybo: but $$\sup_{n}\sin(n)=1$$ still follows from density. What alternative (simpler) approach do you have in your mind? – Jack D'Aurizio Oct 20 '16 at 18:33

I want to calculate the limit: $\displaystyle\limsup_{n\to\infty} \sin(n)$

1 Answers1