Evaluating limits of nested functions

W.l.G. we can assume $x\in[-1,1]$, if $x=0$, the case is easy;

if $x\in(0,1]$, then $\sin x <x$, this implies the sequence{ $x_n=\sin x_{n-1}\}$ is monotone decreasing. Monotone bounded principle implies this sequence has limit $x_0$ and satisfies $$x_0=\sin x_0.$$ This equation has unique solution $x_0=0.$

If $x\in[-1,0)$, you can consider as the same way and get the same limit!