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.