I am trying to show that $\lim_{z\to\infty}f(z)=\lim_{z\to 0}f\left(\frac{1}{z}\right)$ by definition.
My attempt:
Let $\lim_{z\to\infty}f(z)=L$, so by definition, $$\forall\epsilon\in\mathbb{R^+} \ \exists R\in\mathbb{R^+} \ \text{s.t} \ |z|>R\implies |f(z)-L|<\epsilon.$$ Similarly, let $\lim_{z\to 0}f\left(\frac{1}{z}\right)=m$, so by definition, $$\forall\epsilon'\in\mathbb{R^+} \ \exists \delta\in\mathbb{R^+} \ \text{s.t} \ 0<|z|<\delta\implies \left|f\left(\frac{1}{z}\right)-m\right|<\epsilon'$$ But I do not know how to show how they are equal?
You can let $\delta=1/R$ for all $\epsilon' = \epsilon$, and by the uniqueness of limit, $\lim\limits_{z\to 0}f\left(\frac{1}{z}\right)$ is indeed that $L$. (Since you reduced the second limit to the first one)