How find this limit $\lim_{n\to\infty}\underbrace{\sin{\sin{\cdots\sin{x}}}}_{n},x\in R$ [duplicate]

Find this limit $$\lim_{n\to\infty}\underbrace{\sin{\sin{\cdots\sin{x}}}}_{n},x\in R$$

My idea: let $$f(x)=\underbrace{\sin{\sin{\cdots\sin{x}}}}_{n}$$ then $$f(x+2\pi)=f(x)$$,so we only consider $x\in[0,2\pi]$, so define sequence $a_{n}$ such $$a_{1}=\sin{x},a_{n+1}=\sin{a_{n}}$$

so I think we can Discussion of x.can you someone have methods?

other idea: $$|a_{n+2}-a_{n+1}|=|\sin{a_{n+1}}-\sin{a_{n}}|\le |a_{n+1}-a_{n}|$$

$\sin$ is an increasing function in $[-\pi/2,\pi/2]$. The maximum value here is $1$.

Therefore the maximum of $\sin(\sin(x))$ is $\sin(1)$, which is less than $1$, because $\sin(x)\lt x \forall x\gt0$(try to prove by yourself)

Similarly adding another iteration will again decrease the value.

But the value is bounded from below, because with positive input below $\pi/2$, the value will never be negative.

Clearly our required limit is $0$.

You can also show that if you start at the other extreme of $-1$, then it is increasing and bounded above by $0$. Everything else is a subclass of these two cases.