Limits of Reciprocals and Quotients of Complex-Valued Functions

# Limits of Reciprocals and Quotients of Complex-Valued Functions

Let $(S, d_S)$ and $(\mathbb{C}, d)$ be metric spaces where $d$ is the usual metric on $\mathbb{C}$ defined for all $x, y \in \mathbb{C}$ by $d(x, y) = \mid x - y \mid$. Furthermore, let $A \subseteq S$, $p \in S$ be an accumulation point of $A$, $f, g : A \to \mathbb{C}$, and suppose that $\lim_{x \to p} f(x) = a$ and $\lim_{x \to p} g(x) = b$.

Recall from the Limits of Sums and Differences of Complex-Valued Functions page that 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}

Also recall from the Limits of Products of Complex-Valued Functions page that we also have that:

(2)\begin{align} \quad \lim_{x \to p} f(x)g(x) = ab \end{align}

We will now look at the limits of reciprocal and quotients 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 $g : A \to \mathbb{C}$. Suppose that $\lim_{x \to p} g(x) = b$. Then $\lim_{x \to p} \frac{1}{g(x)} = \frac{1}{b}$ provided that $b \neq 0$. |

**Proof:**Let $\epsilon > 0$ be given.

- Since $\lim_{x \to p} g(x) = b$ then for all $\epsilon_1 = \frac{\mid b \mid}{N} \epsilon > 0$ there exists a $\delta_1 > 0$ such that if $d_S(x, p) < \delta$ then:

\begin{align} \quad d(f(x), b) = \mid f(x) - b \mid < \epsilon \quad (*) \end{align}

- Furthermore, since $\lim_{x \to p} g(x) = b \neq 0$ we have that there exists a $\delta_2 > 0$ such that if:

\begin{align} \quad p - \delta_2 < x < p + \delta_2 \end{align}

- We have that $0 \not \in (b - \epsilon, b + \epsilon)$ and that:

\begin{align} \quad b - \epsilon < g(x) < b + \epsilon \end{align}

- We see that if $d_S(x, p) < \delta_2$ then $\mid g(x) \mid > 0$. Furthermore, $\mid g(x) \mid$ is bounded away from $0$ $(p, \delta_2, p + \delta_2)$ and so the $\inf \{ \mid g(x) \mid : d_S(x, p) < \delta_2 \} \neq 0$ so there exists a natural number $N \in \mathbb{N}$ such that $\frac{1}{N} < \mid g(x) \mid$ so $N > \frac{1}{\mid g(x) \mid}$.

- Let $\delta = \min \{ \delta_1, \delta_2 \}$ so that both $(*)$ and $(**)$ hold. Then:

\begin{align} \quad d \left ( \frac{1}{g(x)} , \frac{1}{b} \right ) = \biggr \lvert \frac{1}{g(x)} - \frac{1}{b} \biggr \rvert = \biggr \lvert \frac{b - g(x)}{b g(x)} \biggr \rvert = \frac{\mid b - g(x) \mid}{\mid b \mid \mid g(x) \mid} < {N \epsilon_1}{\mid b \mid} = N \frac{\mid b \mid \epsilon}{N \mid b \mid} = \epsilon \end{align}

- Therefore $\displaystyle{\lim_{x \to p} \frac{1}{g(x)} = \frac{1}{b}}$. $\blacksquare$

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 $\lim_{x \to p} \frac{f(x)}{g(x)} = \frac{a}{b}$ provided that $b \neq 0$. |

**Proof:**From Theorem 1 above we have that $\lim_{x \to p} \frac{1}{g(x)} = \frac{1}{b}$, and so taking the limit of the product of $f(x)$ and $\frac{1}{g(x)}$ we see that:

\begin{align} \quad \lim_{x \to p} \frac{f(x)}{g(x)} = \frac{a}{b} \quad \blacksquare \end{align}