## Proving $\sqrt{n}(x_n)$ converges when $x_n = \sin(x_{n-1}), x_1=1$

This is a problem that showed up on a qual exam that I have been stuck on for a while.

Let $$$$x_n = \sin(x_{n-1}), x_1 = 1$$$$ Prove $$\lim_{n \rightarrow \infty} \sqrt{n} x_n$$ exists and compute its value. The problem gave the following hint: Show that $$$$\frac{1}{x^2_{n+1}} – \frac{1}{x^2_{n}}$$$$ converges to a constant.

I have shown that $$x_n$$ converges to $$0$$, but I am unsure on how to begin bounding $$\sqrt{n}x_n$$ or how the hint is helpful in this problem. I have tried using the MVT to show $$\sin(x_n) \rightarrow 0$$ faster than $$\sqrt{n} \rightarrow \infty$$, but I didn’t get far. Any help will be appreciated.