# 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

• You could instead look at it as the limit $$\lim_{x\to\frac{\pi}{2}} \sin x.$$ This should help some. – Cameron Williams Dec 14 '14 at 19:31
• @CameronWilliams Why can I do such a thing? – The One Dec 14 '14 at 19:40
• It is just a substitution, but it might not be clear to you since Cameron used the same variable. $x\mapsto\frac1{x}$. – slo Dec 14 '14 at 20:36