Limit of Complex Function is Unique
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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
Aiming for a contradiction, suppose $L' \ne L$ 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:
\(\ds \cmod {L - L'}\) | \(=\) | \(\ds \cmod {L - \map f {z^*} + \map f {z^*} - L'}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \cmod {L - \map f {z^*} } + \cmod {\map f {z^*} - L'}\) | Triangle Inequality | |||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon + \epsilon\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \epsilon\) |
This contradicts the choice we made of $\epsilon$.
Hence the result by Proof by Contradiction.
$\blacksquare$
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