# Proving a limit of a trigonometric function: $\lim_{x \to 2/\pi}\lfloor \sin \frac{1}{x} \rfloor=0$

$$\lim_{x \to \frac{2}{\pi}}\lfloor \sin \frac{1}{x} \rfloor=0$$

I need to prove the limit of this using the $\epsilon – \delta$ way but I don’t know how to find $\delta$ when I’m given a trigonometric function I know only how to do it with polynomial functions