EDITED: By Lord Shark the Unknown's answer, seems like the boundary should be $1\over{\vert\sin x\vert}$, but can anyone give a supplement to the proof?
Observing the functions $f_n(x)=\frac{\sin nx}{\sin x}$ for $n\in\mathbb Z$ (Of course for some $a\in $ dom$(f_n)$, it evaluate as $0\over 0$, then they are define as limit of $f(x)$ for $x$ around $a$)
See here for a Desmos graph.
I observe that when $n\to\infty$ , $f_n$ seems to be a 'space filling curve'.
I would like to ask, what is the upper boundary (which is not defined for {$k\pi:k\in\mathbb Z$}), in terms of function?
Should it be $U(x)=\sup${$f_n(x):n\in\mathbb Z$}? I'm not pretty sure about it. (Is it about the irrationality of $\pi$?)
Also I think that it is enough to consider the cases of $f_n$ for $n\in\mathbb N$ since $\forall x\in\mathbb R, \sin(-x)=-\sin(x)$.
Any help will be appreciate. Thank you!
If $x$ is not a rational multiple of $\pi$, then $\sin nx$ will approach $1$ and $-1$ arbitrarily closely as $n$ varies. Therefore $$\sup_n \frac{\sin nx}{\sin x}=\frac1{|\sin x|}$$ as long as $x$ is not a rational multiple of $\pi$.