= Solution
Use the discontinuous example
$$
\boxed{f(x)=\begin{cases}x/2,&0<x\leq1/2,\\(x+1/2)/2,&1/2<x<1.\end{cases}}
$$
In both pieces $0<f(x)<x$. For $x>1/2$, the image stays strictly above $1/2$, and induction gives
$$
f^n(x)=\frac12+2^{-n}\left(x-\frac12\right)\longrightarrow\frac12.
$$
This positive limiting point is not a <fixed point>, since $f(1/2)=1/4$. The discontinuity is precisely why the limit cannot be passed through $f$. Therefore \b[the pointwise zero-limit conclusion can fail without continuity].
Back to article page
