In my book (Churchill), a limit of a function at infinity is defined as: $$ \lim\limits_{z \to \infty}f(z) \equiv \lim\limits_{z \to 0}f\left(\frac{1}{z}\right) $$
But why can't you define the point at infinity as $|z|>\alpha, \forall\alpha\in\mathbb{C}$? So you would get:
$$ \lim\limits_{z \to \infty}f(z) \equiv \lim\limits_{|z| \to \infty}f(z) $$
If this equivalence can be made, could you give me some hints to prove it? (I think it should be something like $\lim\limits_{|r| \to \infty}f(re^{i\theta})$ is independent of $\theta$).
I think this way would be more natural and similar to real analysis.
Your definition works. You could also say that $\lim_{z\to\infty} f(z) = l$ if for every $\epsilon > 0$ there is some compact $K \subset \mathbb{C}$ so that $z\not\in K\implies |f(z) - l| < \epsilon$.