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

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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

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

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