TITLE: Limit of the sequence $(\sin n)^{n}$
QUESTION [10 upvotes]: How to calculate
$$
\lim_{n\to\infty}(\sin n)^{n} \, ?
$$
Is it sufficiently that since $|\sin x|\leq 1$ $\forall x\in\mathbb{R}$ and $|\sin n|<1$ $\forall n\in\mathbb{N}$ then
$$
\lim_{n\to\infty}(\sin n)^{n}=0 \, ?
$$
Is it true that if $|a_{n}|<1$ $\forall n\in\mathbb{N}$ then
$$
\lim_{n\to\infty}(a_{n})^{n}=0 \, ?
$$

REPLY [7 votes]: This limit does not exist, you can find two distinct accumulation points.
By the theorem of Dirichlet, you will find as many integers $p,q$ as you want such that
$$|\pi-\frac pq|<\frac1{q^2},$$or
$$|q\pi-p|<\frac1{q}.$$
Taking the sine and raising to the $p^{th}$ power,
$$|\sin^pp|<\sin^p\frac1{q},$$
i.e. with $q>\frac p4$, $\color{blue}{\sin^pp}$ is arbitrarily close to $\color{blue}0$ infinitely many times.
For the same reason, you will find as many integers $p,q=2^er$ (odd $r$) as you want such that
$$|\frac\pi{2^{e+1}}-\frac pq|<\frac1{q^2},$$or
$$|r\frac\pi{2}-p|<\frac1{q}.$$
Taking the cosine (a decreasing function) and raising to the $p^{th}$ power, $$|\sin^pp|>\cos^p\frac1{q},$$
i.e. noting that $q>2^{e-1}p$,
 $\color{blue}{\sin^pp}$ is arbitrarily close to $\color{blue}1$ infinitely many times (as $\cos\frac1q=1-o\left(\frac1q\right)$).