2.3

2.3 Prove that if a limit exists, then it is unique.
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Proof: Contradiction
Assume two limits exist, call them \(lim_{z \rightarrow z_0}f(z)=a\) and \(lim_{z \rightarrow z_0}f(z)=b\), such that \(a \neq b\). Choose \(\epsilon =\frac{|a-b|}{2}\). Then by definition of limit, there exists \(\delta_a\) such that \(|z-z_0| < \delta_a \Rightarrow |f(z)-a|< \epsilon\), and there exists \(\delta_b\) such that \(|z-z_0| < \delta_b \Rightarrow |f(z)-a|< \epsilon\). Now choose \(\delta=min\lbrace \delta_a,\delta_b \rbrace\), thus both \(|z-z_0| < \delta_a\) and \(|z-z_0| < \delta_b\) are true. Now consider \(|a-b|\): \[\begin{aligned} |a-b|&=|a-f(z)+f(z)-b| &\text{ by algebra}\\ &\leq|a-f(z)|+|f(z)-b| &\text{ by triangle inequality}\end{aligned}\] From above, we know \(|f(z)-a|< \epsilon\) and \(|f(z)-a|< \epsilon\), thus \(|a-b|<2\epsilon\), ie \(\epsilon<\frac{|a-b|}{2}\); which is a contradiction since we chose \(\epsilon =\frac{|a-b|}{2}\). Thus \(a=b\) which implies \(lim_{z \rightarrow z_0}f(z)=a\) is equal to \(lim_{z \rightarrow z_0}f(z)=b\), therefore, if a limit exists, then it is unique.