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Starting from $0<a_1\le b_1$, replace the lower endpoint by the geometric mean and the upper endpoint by the arithmetic mean. The arithmetic-geometric mean inequality gives $a_n\le a_{n+1}\le b_{n+1}\le b_n$. The bounded monotone sequence theorem gives limits, and the arithmetic recurrence forces them to coincide. The limiting value is the arithmetic-geometric mean of the initial pair.

Fix $x\in (0,1)$ and put $x_0=x$, $x_{n+1}=f(x_n)$. Then $0<x_{n+1}<x_n\le x_0<1$. The bounded monotone sequence theorem gives a limit $\ell \in [0,x_0]$. If $\ell >0$, it lies in the domain, where continuity allows passage to the limit in the recurrence: $\ell =f(\ell )$. That is a fixed point, contrary to $f(\ell )<\ell$. Hence $\boxed{f^n(x)\to 0\text{ for every }x\in (0,1)}$. Continuity is used at a possible positive limiting point, not at the excluded endpoint zero.

No: there need not be even one orbit tending to zero. Partition the domain into the disjoint intervalsand, for the unique $j$ with $x\in I_j$, defineThe left endpoint is strictly below $x$, so $0<f(x)<x$. Moreover $f(x)>1/(j+1)$ and $f(x)\le x\le 1/j$, so $f(x)$ remains in the same interval. For every iterate,Every starting point belongs to some finite-index interval, so this proves the claim for all $x\in (0,1)$. The open left endpoints matter: an orbit approaches a boundary but never crosses it. This is discontinuous trapping of decreasing iterates, not a violation of the bounded monotone sequence theorem; each orbit does converge, just to a positive value where continuity fails.

The arithmetic-geometric mean inequality givesBy induction these inequalities hold at every step, so $(a_n)$ is increasing and bounded above by $b$, while $(b_n)$ is decreasing and bounded below by $a$. The bounded monotone sequence theorem gives limits $\alpha ,\beta$ with $0<a\le \alpha \le \beta \le b$. Taking limits in the arithmetic mean recurrence gives $\beta =(\alpha +\beta )/2$, whence $\boxed{\alpha =\beta }$. Thus both sequences converge to the same positive limit, the arithmetic-geometric mean iteration's limit. No elementary closed formula for that limit is needed.