I am reading an introductory book on complex analysis. They define the neighborhood of infinity as $|z| > 1/\epsilon$ for some $\epsilon >0$. Then they wish to show the following statement: $$ \lim_{z \rightarrow \infty}f(z) = w_0 \quad \text{iff} \quad \lim_{z \rightarrow 0}=f\left(\frac{1}{z}\right)$$
Their proof goes as follows:
In the forward direction, note by definition we have that $\forall \epsilon > 0 \, \exists \delta > 0$ st: $$|z| > 1/\delta \implies |f(z) - w_0| < \epsilon$$ Replacing $z$ by $1/z$ we obtain: $$0 < |z - 0| > \delta \implies |f\left(\frac{1}{z}\right) - w_0| < \epsilon$$
Why can they just "replace" $z$ with $1/z$? Can I do this with any function of $z$ like $z^2$?
You can substitute anything you like but you must respect the inequalities.
In this case we have that $|z| > 1/\delta \implies |f(z) - w_0| < \epsilon$ and a straight replacement gives us $|\frac{1}{z}| > 1/\delta \implies |f(\frac{1}{z}) - w_0| < \epsilon$.
Taking reciprocals on the left side of the implication, we get $|z|\lt\delta\implies |f(\frac{1}{z})-w_0|\lt\epsilon$