Limit when $n\rightarrow\infty$ of $\text{sgn}(\sin(2^n \pi x))$ with $x\in(0,1)$ fixed. I am working on determining the limit of the sign function of that expression above. My conjecture is that, for a fixed value of $x$, the limit does not exist, but I can't seem to be able to prove it. Is there a formal way to show that the sign function varies on every point in the domain? 
 A: Given a real $a$, let $\{a\}$ denote the fractional part of $a$
Notice that $\operatorname{sgn}(\sin(2^n\pi x))$ depends only on the remainder of the division of $2^n\pi x$ by $4\pi$. More specifically, take an integer $k$ such that $2k\pi\leq 2^n\pi x<2(k+1)\pi$. Then $k\leq 2^{n-1}x<k+1$, so the sign of $2^n\pi x$ depends only on $F_n:=\{2^{n-1}x\}$: if $F_n<1/2$, we have sign $\operatorname{sgn}(\sin(2^n\pi x))=1$. If $F_n>1/2$, we have $\operatorname{sgn}(\sin(2^n\pi x))=-1$. If $F_n=1/2$ we have $\operatorname{sgn}(\sin(2^n\pi x))=0$
So, let's analyse what happens to $F_n$ for distinct values of $x$:

*

*If $x=m2^{-k}$ for some $m$ and $k$ in $\mathbb{N}$, $2^n\pi x=2^{n-m}\pi$, so $\operatorname{sgn}(\sin(2^n\pi x))=0$ for $n>k$, hence $\operatorname{sgn}(\sin(2^n\pi x))\to 0$ as $n\to \infty$.


*Now, suppose that $x\neq m 2^{-k}$ for any choice of $m$ and $k$ in $\mathbb{N}$.
You should be convinced that the following formula is valid: $F_{n+1}=\{2F_n\}$. Using this, it is easy to show (by induction, if you want) that $F_n\neq m2^{-k}$ for any choice of $m$ and $k\in\mathbb{N}$ as well (so in particular, $F_n\neq 1/2$ for all $n$).
Pick any $n_0\in\mathbb{N}$, and say, for example, that $0<F_{n_0}<1/2$. The formula $F_{n+1}=\{2F_n\}$ implies that the distance from $F_n,F_{n+1},\ldots$ to $0$ will be always increasing, and it will eventually be bigger than $1/2$, say $F_{n_1}>1/2$ (with $n_1>n_0$). But that same formula then implies that the distance from $F_{n_1},F_{n_1+1},\ldots$ to $1$ will also increase, eventually getting bigger than $1/2$, say $F_{n_2}<1/2$. If we proceed in this way, we find a sequence of naturals $n_0<n_1<n_2<\cdots$ such that $0<F_{n_p}<1/2$ for $p$ odd and $1/2<F_{n_p}<1$ for $p$ even (if we had started with $F_{n_0}>1/2$, a similar argument would be valid). Therefore, we have the subsequence $\operatorname{sgn}(\sin(2^{n_p}\pi x))=(-1)^{p+1}$, which does not converge, so the whole sequence does not converge as well.
Concluding, $\operatorname{sgn}(\sin(2^n\pi x))$ converges iff $x=m2^{-k}$ for some $m$ and $k$ in $\mathbb{N}$, and in this case it converges to $0$.
A: *

*If $x\in\mathbb Q$:


*

*If $2^{k}\mid x,k\in\mathbb Z$. limit is zero because n can always become more large than k even if k is negative, so we get an integral multiple of $\pi$, so limit becomes zero.

*If $2^{k}\nmid x,k\in\mathbb Z$. limit is not defined in the similiar way $\lim_{n\to\infty}\sin n$ is undefined.


*If $x\not\in\mathbb Q$, similar to previous case.
A: If $x$ is not a dyadic rational, i.e. not of the form $m/2^k$ for some positive integers $m,k$, it  has a non-terminating base-$2$ expansion $0.d_1 d_2 d_3 \ldots$.  Then $\sin(2^n \pi x) > 0$ iff the integer part of $2^n x$
is even, i.e. iff $d_n = 0$, and $ < 0$ otherwise.  Since there are infinitely many $0$'s and infinitely many $1$'s, the limit does not exist. 
If $x$ is a dyadic rational, $\sin(2^n \pi x) = 0$ for sufficiently large $n$.   
