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