Limits of Products of Complex-Valued Functions

Limits of Products of Complex-Valued Functions

Recall from the Limits of Sums and Differences of Complex-Valued Functions page that if $(S, d_S)$ and $(\mathbb{C}, d)$ are metric spaces where $d$ is the usual metric on $\mathbb{C}$ defined for all $x, y \in \mathbb{C}$ by $d(x, y) = \mid x - y \mid$ then if $A \subseteq S$, $p \in S$ is an accumulation point of $A$, and $f, g : A \to \mathbb{C}$ are such that $\lim_{x \to p} f(x) = a$ and $\lim_{x \to p} g(x) = b$ then:

(1)
\begin{align} \quad \lim_{x \to p} [f(x) + g(x)] = a + b \quad \mathrm{and} \quad \lim_{x \to p} [f(x) - g(x)] = a - b \end{align}

We will now look at an important result regarding the limits of products of complex-valued functions.

Theorem 1: Let $(S, d_S)$ and $(\mathbb{C}, d)$ be metric spaces where $d$ is the usual metric defined for all $x, y \in \mathbb{C}$ by $d(x, y) = \mid x - y \mid$. Also let $A \subseteq S$, $p \in S$ be an accumulation point of $A$, and $f, g : A \to \mathbb{C}$. Suppose that $\lim_{x \to p} f(x) = a$ and $\lim_{x \to p} g(x) = b$. Then $\lim_{x \to p} f(x)g(x) = ab$.
  • Proof: Let $\epsilon$ be such that $0 < \epsilon < 1$. If $\epsilon \geq 1$ then we can such take a smaller $\delta$ corresponding to a smaller $\epsilon_0$ such that $0 < \epsilon_0 < 1$ and the inequalities will still hold.
  • Since $\lim_{x \to p} f(x) = a$ we have that for $\epsilon_1 = \frac{\epsilon}{\mid a \mid + \mid b \mid + 1} > 0$ there exists a $\delta_1 > 0$ such that if $x \in D(f) \setminus \{ p \}$ and $d_S(x, p) < \delta_1$ then $d(f(x), a) < \epsilon_1$, $(*)$.
  • Similarly, since $\lim_{x \to p} g(x) = b$ we have that for $\epsilon_2 = \frac{\epsilon}{\mid a \mid + \mid b \mid + 1}$ there exists a $\delta_2 > 0$ such that if $x \in D(g) \setminus \{ p \}$ and $d_S(x, p) < \delta_2$ then $d(g(x), b) < \epsilon_2$, $(**)$.
  • Now since $\epsilon < 1$ we have that $\epsilon_1, \epsilon_2 < 1$, so then:
(2)
\begin{align} \quad \mid f(x) \mid = \mid a + f(x) - a \mid \leq \mid a \mid + \mid f(x) - a \mid < \mid a \mid + \epsilon_1 < \mid a \mid + 1 \quad (***) \end{align}
  • Let $\delta = \min \{ \delta_1, \delta_2 \}$. Then if $x \in D(f) \setminus \{ p \}$ and $d_S(x, p) < \delta$ then $d_S(x, p) < \delta_1$ AND $d_S(x, p) < \delta_2$ so $(*)$, $(**)$, and $(***)$ will hold and:
(3)
\begin{align} \quad d(f(x)g(x), ab) & = \mid f(x)g(x) - ab \mid \\ \quad &= \mid f(x)g(x) - bf(x) + bf(x) -ab \mid \\ \quad &= \mid f(x)[g(x) - b] + b[f(x) - a]\mid \\ \quad & \leq \mid f(x) \mid \mid g(x) - b \mid + \mid b \mid \mid f(x) - a \mid \\ \quad & < [\mid a \mid + 1] \epsilon_2 + \mid b \mid \epsilon_1 \\ \quad & < [\mid a \mid + 1] \frac{\epsilon}{\mid a \mid + \mid b \mid + 1} + \mid b \mid \frac{\epsilon}{\mid a \mid + \mid b \mid + 1} \\ \quad & < [\mid a \mid + \mid b \mid + 1] \frac{\epsilon}{\mid a \mid + \mid b \mid + 1} \\ \quad & < \epsilon \end{align}
  • Therefore $\lim_{x \to p} f(x)g(x) = ab$. $\blacksquare$
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License