# Limit of Function Unique

## Theorem

### Complex Analysis

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

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

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

Then that limit $L$ is unique.

## Proof

### Proof for Complex Analysis

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

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

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

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

Because $z_0$ is a limit point:

$\exists z^* \in S: 0 < \left\vert{z - z_0}\right\vert < \min \left\{{\delta_1, \delta_2}\right\}$

Then:

 $\displaystyle \left\vert{L - L'}\right\vert$ $=$ $\displaystyle \left\vert{L - f \left({z^*}\right) + f \left({z^*}\right) - L'}\right\vert$ $\displaystyle$ $\le$ $\displaystyle \left\vert{L - f \left({z^*}\right)}\right\vert + \left\vert{f \left({z^*}\right) - L'}\right\vert$ Triangle Inequality $\displaystyle$ $<$ $\displaystyle \epsilon + \epsilon$ $\displaystyle$ $=$ $\displaystyle 2 \epsilon$

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

$\blacksquare$