Combination Theorem for Limits of Functions/Complex/Product Rule

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Theorem

Let $\C$ denote the complex numbers.

Let $f$ and $g$ be complex functions defined on an open subset $S \subseteq \C$, except possibly at the point $c \in S$.

Let $f$ and $g$ tend to the following limits:

$\ds \lim_{z \mathop \to c} \map f z = l$
$\ds \lim_{z \mathop \to c} \map g z = m$


Then:

$\ds \lim_{z \mathop \to c} \ \paren {\map f z \map g z} = l m$


Proof

Let $\sequence {z_n}$ be a sequence of elements of $S$ such that:

$\forall n \in \N: z_n \ne c$
$\ds \lim_{n \mathop \to \infty} z_n = c$


By Limit of Complex Function by Convergent Sequences:

$\ds \lim_{n \mathop \to \infty} \map f {z_n} = l$
$\ds \lim_{n \mathop \to \infty} \map g {z_n} = m$


By the Product Rule for Complex Sequences:

$\ds \lim_{n \mathop \to \infty} \paren {\map f {z_n} \map g {z_n} } = l m$


Applying Limit of Complex Function by Convergent Sequences again, we get:

$\ds \lim_{z \mathop \to c} \paren {\map f z \map g z} = l m$

$\blacksquare$