Limit of Complex Function is Unique

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

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

Suppose that $$\lim_{z \to z_0} f \left({z}\right) = L$$.

Then that limit $$L$$ is unique.

Proof for Complex Analysis
Suppose $$L' \ne L$$ is another limit of $$f \left({z}\right)$$ at $$z_0$$.

Let us take $$\epsilon = \frac {\left|{L - L'}\right|} 2$$.

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


 * $$z \in S, 0 < \left|{z - z_0}\right| < \delta_1 \implies \left|{f \left({z}\right) - L}\right| < \epsilon$$;
 * $$z \in S, 0 < \left|{z - z_0}\right| < \delta_2 \implies \left|{f \left({z}\right) - L'}\right| < \epsilon$$;

Because $$z+0$$ is a limit point, $$\exists z^* \in S: 0 < \left|{z - z_0}\right| < \min \left\{{\delta_1, \delta_2}\right\}$$.

Then:

$$ $$ $$ $$

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