Limit computation $\lim_{n \to \infty } \left |\sin n \right |n=\infty $ Does the 
$\lim_{n \to \infty } \left |\sin n  \right |n=\infty $
Thank you!
 A: Since $\pi$ is irrational, there are infinitely many rational numbers $p/q$ such that 
 $$ \left|\pi-\frac pq\right|<\frac1{q^2}. $$
This gives us that $|p-q\pi|<\frac1q$. Hence
 $$ |\sin(p)|=|\sin(p-q\pi+q\pi)|=|\sin(p-q\pi)|<\sin(1/q)<1/q, $$
and $p|\sin(p)|<2\pi$. This holds for arbitrarily large values of $p$, so certainly the sequence $(n|\sin n|)_{n\ge1}$ does not diverge to infinity.

On the other hand, $(\sin n)_{n\ge1}$ is dense in $[-1,1]$, which follows from a careful application of the pigeonhole principle. What matters here is that there are arbitrarily large $n$ such that $\sin n>1/2$, so there is a subsequence of $(n|\sin n|)_{n\ge1}$ that diverges to infinity. In particular, the sequence itself does not converge. A more delicate question, which seems open, is whether $(n\sin n)_{n\ge1}$ is dense in $\mathbb R$. And, sure enough, this has been asked here before.
To address a recent comment to the question: Note that replacing $n\in\mathbb N$ with a real variable $x\in\mathbb R_{>0}$ makes the problem uninteresting: There are arbitrarily large $x$ such that $\sin x=1$ and arbitrarily large $x$ such that $\sin x=-1$. By continuity, any real number $t$ has the form $x\sin x$ for infinitely many values of $x$ (that can be taken to be arbitrarily large.)
