Does $\lim\limits_{n\to+\infty}|\sin{2^n}|^{\frac1n}$ exist? More generally, for what real number x does $\displaystyle\lim_{n\to+\infty}|\sin2^nx|^\frac1n$ exist?
Actually it's already known that if you replace the $2^n$ with $n$, then the answer is 1. That is, $$\lim\limits_{n\to+\infty}|\sin n|^\frac1n=1$$.
But when it seems that I could not get the same information about how $\frac{2^n} m$ goes to $\pi$.
To answer this question, one has to consider two things:

*

*$\liminf\limits_{n\to+\infty}|2^n-m\pi|$ $=$ ?

*If the answer to 1) is $0$, then how fast does the infimum actually come to $0$?

 A: Just a few elementary facts that you didn't mention.
Let $f(t)= t-\lfloor t +1/2\rfloor$. Note that $f(2t) = f(2f(t))$, thus for $f(t)\in (-1/4,1/4)$ we have $f(2t)=2f(t)$.

*

*If $|\sin(2^n \pi x)|^{1/n}\to 0$ then $f(2^nx)\to 0$.
Thus for all $n\ge N$ large enough we have $f(2^n x)\in (-1/4,1/4)$ so that $f(2^{n+1} x)=2 f(2^n x)$ and hence $f(2^n x) = 2^{n-N} f(2^N x)$ which implies that $f(2^N x)=0$ ie.  $x\in \Bbb{Z}[2^{-1}]$.


*Next assume that $|\sin(2^n \pi x)|^{1/n}\to r\in (0,1)$, thus $|\sin(\pi f(2^n  x))|= |\sin(2^n \pi x)| \sim r^n, \pi f(2^n  x) \sim r^n$ so that $\pi f(2^{n+1}  x) \sim 2 r^n$ and $r^{n+1} \sim 2 r^n$ is a contradiction.


*Whence for $x\not\in \Bbb{Z}[2^{-1}]$, if $\lim_{n\to \infty}|\sin(2^n \pi x)|^{1/n}$ exists then the limit is $1$.


*If $x\in \Bbb{Q}-\Bbb{Z}[2^{-1}]$ then the binary expansion of $x$ is infinite periodic so that $f(2^n x)$ is periodic and $\lim_{n\to \infty}|\sin(2^n \pi x)|^{1/n}=1$.
With things like $x=\sum_{m\ge 1} 2^{-m!}$ then $\lim_{n\to \infty}|\sin(2^n \pi x)|^{1/n}$ doesn't exist.
For $x\not \in\Bbb{Z}[2^{-1}]$ let $c_l$ be the indice of the first $\underbrace{0\ldots 0}_l$ or $\underbrace{1\ldots 1}_l$ motif found in the binary expansion of $x$ (if it doesn't exist set $c_l=\infty$). Then $\lim_{n\to \infty}|\sin(2^n \pi x)|^{1/n}$ exists (thus is $=1$) iff $c_l/l\to \infty$.
