Certainly i dont understand how that step is done,

in fact it would seem to me that \(\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)|\) doesnt make any sense since the left side does not depend on x.

Returning to the proof, earlier we had that

\(\displaystyle \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|<2\epsilon x\) for all n if \(\displaystyle 0< x <\alpha\)

Taking \(\displaystyle n\to\infty\) and using the continuity of the abs value we get that since if \(\displaystyle n\to \infty\) then \(\displaystyle y=\frac{x}{2^n}\to 0\),

in our case :

if \(\displaystyle n\to \infty \) then \(\displaystyle f(\frac{x}{2^n})=f(y)\to 0\).

And we end up having \(\displaystyle |f(x)|\leq 2\epsilon x \) from where the proof follows quite easily.