How to prove sin(sin(sin...(sin(1))) converges to 0 without using continuity?

Question

Using continuity I was able to show the sequence $x_0 = 1$, $x_{n+1} = sin(x_n)$ converges to 0, but I was wondering if there was a way to prove it using only properties and theorems related to sequences and series, without using continuity.

So far, I know the sequence is monotonically decreasing and bounded below by 0, so it must converge to its infimum. From here I'm not exactly sure how to show 0 is the infimum of this set of numbers.

Alternatively, I could check convergence to $0$ by comparison, but no sequences come to mind that are greater than the given sequence for all $n \geq N$ and converge to $0$.

I've already seen the answers at the following:

Compute $ \lim\limits_{n \to \infty }\sin \sin \dots\sin n$

Prove that $\sin(\sin...(\sin(x))..)$ converges asymptotically to zero

Answer 1

We know that for all $x \in (0,1]$, we have $$ 0 < \sin x < x $$ From there, you can show that $x_0,x_1,\dots$ is a strictly monotonically decreasing sequence. Is this enough? That is, can we forgo continuity? No. As a counterexample, consider the seuqence $x_k = f(x_{k-1}); x_0 =1$ with $$ f(x) = \begin{cases} \sin(x) + 0.1 & x^*<0\leq 1\\ \sin(x) & 0 < x \leq x^* \end{cases} $$ Where $x^*$ is the positive solution to $= x^* = \sin(x^*) + 0.1$. We note that $x_k$ is a strictly monotonically decreasing sequence converging to $x^*$ (and therefore bounded below by $0$), and we even have $f(x)<x$ for every point in $(0,1]$. However, because of the lack of continuity, this is not enough to "force" $\{x_k\}$ to converge to $0$.

Answer 2

$$\sin(\sin...sin(1)))....)) =L$$ $$ \sin(\sin...sin(1)))....)) =arcsin(L)$$ $$L=arcsin(L)$$ $$\sin(L) =L$$ $$L=0$$