Could someone explain how to correctly prove that $\lim_{n\to\infty}\sin\frac{1}{n}$ where $n=1,2,\cdots,n$ doesn't exist. I have no problem with it if $\sin\frac{1}{x}$ where $x$ is real, because just taking values $x=\frac{2}{(2n-1)\pi}, x=\frac{1}{n\pi}, x=\frac{2}{(2n+1)\pi}$ it is clear, for example, by Cauchy criterion. But what about when $n$ takes natural values?
This is posterior edit. It corresponds to comments and the answers I got till now: I am pretty sure that this limit doesn't exist. The graph of this function goes from $-1$ to $0$ to $\,1$ and so on infinitely many times as $\frac{1}{n}$ approaches $0$. So there is no way you could find $n$ big enough so that $|\sin \frac{1}{n}-0|<\epsilon$ if $\epsilon<1$