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In my book (Churchill), a limit of a function at infinity is defined as: $$ \lim\limits_{z \to \infty}f(z) \equiv \lim\limits_{z \to 0}f(\frac{1}{z}) $$

But why can't you define the point at infinty 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.

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The two are equivalent. $\lvert z\rvert > \alpha \iff \left\lvert \frac1z\right\rvert < \frac1\alpha$ (for $\alpha > 0$). –  Daniel Fischer Oct 7 '13 at 13:41
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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 $z\not\in K\implies |f(z) - l| < \epsilon$.

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How can a prove that $ \lim\limits_{|r| \to \infty}f(re^{i\theta})=l $ is independent of $\theta$ so both definitions are equivalent? –  jinawee Oct 11 '13 at 9:20
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