Let
$$ f(x)=1-\frac{1}{x\lvert x\rvert_2}. $$
Show that $f^m(x)$ converges to $0$ for all $x\in\mathbb{N_{>0}}$, for sufficiently high $m\in\mathbb{N}$.
In a nutshell, $x\lvert x\rvert_2$ boils down to the odd factors of $x$. $\lvert x\rvert_2$ is the $2$-adic metric of $x$, defined by $\lvert x\rvert_2=\frac{1}{2^p}$, where $x=2^p\cdot\frac{r}{q}$ and $r,q$ are odd numbers. Note that the question is, whether for an initial integer input $x$, $f^m(x)$ converges, however $f(x)$ must be defined over rationals so we have
$$f(x)=1-\frac{1}{x\lvert x\rvert_2}\quad \mathbb{Q}\mapsto\mathbb{Q}.$$
Let $x_{m+1}=f(x_m)$. Show that $\forall x_0\in\mathbb{N_{>0}}\exists > n\mid (f^m(x)=0\forall m\geq n)$
UPDATE
I'm currently investigating whether Mahler's theorem and Newton's forward difference formula have something to say. Forward difference formula looks promising on the face of it but I haven't studied that in depth.