Limit of Complex Function is Unique

Theorem
Let $f: S \to \C$ be a complex function.

Let $z_0$ be a limit point of $S$.

Suppose that $\ds \lim_{z \mathop \to z_0} \map f z = L$.

Then that limit $L$ is unique.

Proof
$L' \ne L$ is another limit point of $\map f z$ at $z_0$.

Let us take $\epsilon = \dfrac {\cmod {L - L'} } 2$.

Then we can find $\delta_1 > 0, \delta_2 > 0$ such that:


 * $z \in S, 0 < \cmod {z - z_0} < \delta_1 \implies \cmod {\map f z - L} < \epsilon$
 * $z \in S, 0 < \cmod {z - z_0} < \delta_2 \implies \cmod {\map f z - L'} < \epsilon$

Because $z_0$ is a limit point:
 * $\exists z^* \in S: 0 < \cmod {z - z_0} < \min \set {\delta_1, \delta_2}$

Then:

This contradicts the choice we made of $\epsilon$.

Hence the result by Proof by Contradiction.