= Solution
Fix $x\in(0,1)$ and put $x_0=x$, $x_{n+1}=f(x_n)$. Then $0<x_{n+1}<x_n\leq x_0<1$. 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 \b[$\boxed{f^n(x)\to0\text{ for every }x\in(0,1)}$]. Continuity is used at a possible positive limiting point, not at the excluded endpoint zero.
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