# Proof of Quotient Rule for Sequences

$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$
Quotient Rule
Suppose that $(a_n)$ and $(b_n)$ are two convergent sequences with $a_n\to a$ and $b_n\to b$. If $b_n\neq 0$ for all $n\in \N$ and $b\neq 0$, then $a_n / b_n \to a/b$.

First we need a lemma.
Lemma
Suppose that $b_n\to b$ with $b_n\neq 0$ for all $n\in \N$ and $b\neq 0$. Then, there exists $m\in \R$ such that $0\lt m\leq |b_n|$ for all $n\in \N$.

Proof
As $b_n\to b$, (taking $\varepsilon = |b|/2$), there exists $N$ such that $|b_n-b|<|b|/2$ for all $n>N$.
But then, by the reverse triangle inequality,
$|b_n|=|b-(b-b_n)|\geq |b|-|b-b_n|\geq |b|/2.$
Let
$m=\min \{ |b_1| ,|b_2| , |b_3| , \dots , |b_{N}|, |b|/2 \} .$
Then $m$ is such that $m\neq 0$ and $m\leq |b_n|$ for all $n\in \N$.

Proof
This will follow from the Product Rule if we can prove that if $b_n\neq 0$ and $b\neq 0$, then $1/b_n\to1/b$.

From the assumptions we know from the preceding lemma that there exists
$m\in \R$ such that $0\lt m\leq |b_n|$ for all $n\in \N$.

Given $\varepsilon>0$, there exists $N$ such that $|b_n-b|\lt m|b|\varepsilon$ for all $n>N$.
Then, for all $n>N$,
$\left| \frac{1}{b_n} -\frac{1}{b} \right| = \frac{|b-b_n|}{|b_n||b|} \leq \frac{|b_n-b_|}{m|b|} \lt \varepsilon .$