Choose $x_0\in (0,\pi)$ arbitrarily, and define a sequence $x_n$ iteratively by $x_n=\sin(x_{n-1})$ for each $n\ge 1$. Show that $\lim_{n\to\infty} \sqrt{n}\cdot x_n =\sqrt{3}$.

I am not able to solve this. Not even a single idea which can help came to my mind. So any hint or solution.

  • $\begingroup$ I think you should use square brackets for the range of numbers $x_0$ can start from. Because I don't think $x_0$ = 0 or $\pi$ works. Maybe it's $x_0 \in \, ]0, \pi[$? $\endgroup$
    – Shuri2060
    Aug 17, 2016 at 18:38
  • $\begingroup$ Hmmm, that's true..but any idea $\endgroup$
    Aug 17, 2016 at 18:39
  • $\begingroup$ @QuestionAsker Usual brackets have exactly this purpose. $x_0 \neq 0, \pi$. $\endgroup$
    – Crostul
    Aug 17, 2016 at 18:40
  • $\begingroup$ @Crostul My bad - am unfamiliar with that $\endgroup$
    – Shuri2060
    Aug 17, 2016 at 18:40
  • $\begingroup$ @QuestionAsker no, it's correct as it stands, since $[0,\pi]$ would mean you include these two points, $(0,\pi)$ means you don't $\endgroup$
    – user190080
    Aug 17, 2016 at 18:40