Clarification: Proof of the quotient rule for sequences

Question

My Problem I am currently looking for a proof for the quotient rule for sequences:

$a_n$ and $b_n$ are two sequences with the limes a,b. So:

When

$ a_n \rightarrow a$ and $ b_n \rightarrow b$

Then:

$\frac {a_n}{b_n} \rightarrow \frac{a}{b}$

Awesome stuff, but how do i prove it?

Solving attempts

I found a proof in my textbook, but i have a hard time understanding it. It goes as follows.

To prove the quotient rule, we have to acknowledge: $b \ne 0$.Then

$b_n \rightarrow b > 0 $

Thus, for every $\epsilon : \frac{|b|}{2}$, There is a $n_0$, so that $|b_n|>|b|-\epsilon= \frac{|b|}{2}$ for every $n>n_0$. For those n's (This is the thing i don't get) You can say:

$$| \frac {1}{b_n}-\frac {1}{b}|=|\frac{b-b_n}{b_nb}| \le \frac {2}{|b|^2}|b-b_n|$$

Because of the factor rule(Which is the equivalent of the quotient rule, just for factors), the right side of the inequality goes towards 0. Because of the rule 22.3 you can conclude that $\frac{1}{b_n} \rightarrow \frac{1}{b}$, and if you apply the product rule again, also that $\frac{a_n}{b_n} \rightarrow \frac {a}{b}$.

Rule 22.3:

$\alpha_n $is a null sequence. If the inequality $|a_n- a|\le \alpha_n$ is valid from a certain point with a limited number of exeptions, then $\alpha_n \rightarrow n$

I understand the beginning of the proof, but not how we get to the right side of the inequality. And i mean i understand why this thing on the right is a null sequence. But how does this prove our point.

If somebody could clarify or point me to another proof of this rule, that is maybe easier to understand, i would be very grateful.

Answer 1

If, for eny $\epsilon > 0 \in \mathbb{R}$, there exists an $n \in \mathbb{Z}$ such that $\forall i > n, |a_i - c| < \epsilon $, then $c$ is defined as the limit of the sequence of $a_i$.

Assuming the limit of sequence $a_i$ is $a > 0$, and the limit of $b_i$ is $b > 0$, we want to prove the limit of $\dfrac{a_i}{b_i} = \dfrac{a}{b}$.

If we apply the limit definitions to $a_i$ and $b_i$ at $\delta < \frac{b}{2}$, how different can $\dfrac{a_i}{b_i}$ be from $\dfrac{a}{b}$?

We apply the limit definition and get a $n_a$ such that $\forall i \ge n_a, |a_i - a| < \delta$, and we get a $n_b$ such that $\forall i \ge n_b, |b_i - b| < \delta$. Let $n' = \max(n_a, n_b)$. We know that for all $i > n', $ both $|a_i - a| < \delta$ and $|b_i - b| < \delta$. We can also say this as $a - \delta \le a_i \le a + \delta$ and $b - \delta \le b_i \le b + \delta$.

So $\dfrac{a - \delta}{b + \delta} \le \dfrac{a_i}{b_i} \le \dfrac{a+\delta}{b-\delta}$. How far do these minimum and maximum bounds stray from $\dfrac{a}{b}$?

$$\begin{align} \frac{a}{b} - \frac{a-\delta}{b+\delta} & = \frac{a(b+\delta) - (a-\delta)b}{b(b+\delta)}\\ & = \frac{ab + a\delta - ab + \delta b}{b(b+\delta)} \\ & = \frac{(a+b)\delta}{b(b+\delta)}\\ & < \frac{a+b}{b^2}\delta \end{align}$$

$$\begin{align} \frac{a + \delta}{b - \delta} - \frac{a}{b} & = \frac{(a + \delta)b - a(b - \delta)}{b(b - \delta)}\\ & = \frac{ab + \delta b - ab + a\delta}{b(b - \delta)}\\ & = \frac{a + b}{b(b - \delta)}\delta\\ & < \frac{a + b}{b\frac{b}{2}}\delta && \text{remember $\delta < \frac{b}{2}$}\\ \frac{a + \delta}{b - \delta} - \frac{a}{b} & < \frac{a + b}{b^2}2\delta \end{align} $$

So now we can prove the limit of $\dfrac{a_i}{b_i}$. When we are given an $\epsilon > 0$, we can use the limits of $a_i$ and $b_i$ for the convenient $\delta = \min(\frac{b}{2}, \frac{b^2}{2(a+b)}\epsilon)$ to get $n_a$ and $n_b$. Then we determine $n' = \max(n_a,n_b)$. Then (omitting the trivial $\frac{b}{2} < \frac{b^2}{2(a+b)}\epsilon $ case) we know that for all $i > n'$,

$$\begin{align} |\frac{a_i}{b_i} - \frac{a}{b}| & \le \frac{2(a+b)}{b^2}\delta \\ & \le \frac{2(a+b)}{b^2}\frac{b^2}{2(a+b)}\epsilon \\ & \le \epsilon \end{align}$$

And so we fulfill the definition of the limit.