Combination Theorem for Limits of Mappings/Metric Space/Quotient Rule

Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $\R$ denote the real numbers.

Let $f: A \to \R$ and $g: A \to \R$ be real-valued functions defined on $A$, except possibly at the point $a \in A$.

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

$\ds \lim_{x \mathop \to a} \map f x = l$

$\ds \lim_{x \mathop \to a} \map g x = m$

Then:

$\ds \lim_{x \mathop \to a} \frac {\map f x} {\map g x} = \frac l m$

provided that $m \ne 0$.

Proof

Let $\sequence {x_n}$ be any sequence of elements of $S$ such that:

$\forall n \in \N_{>0}: x_n \ne a$

$\ds \lim_{n \mathop \to \infty} \ x_n = a$

By Limit of Function by Convergent Sequences:

$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$

$\ds \lim_{n \mathop \to \infty} \map g {x_n} = m$

By the Quotient Rule for Real Sequences:

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

provided that $m \ne 0$.

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

$\ds \lim_{x \mathop \to a} \frac {\map f x} {\map g x} = \frac l m$

provided that $m \ne 0$.

$\blacksquare$