# Proof of quotient rule for limits, is it correct?

PROPOSITION *Let $$(a_n)_{n \in \mathbb{N}}$$ and $$(b_n)_{n \in \mathbb{N}}$$ be convergent sequences, with $$\lim_{n \to \infty}a_n = a , \: \lim_{n \to \infty}b_n = b$$ then $$\lim_{n \to \infty}\frac{a_n}{b_n} = \frac{a}{b}$$ PROOF

Let $$\varepsilon > 0$$ be given and choose$$K \in \mathbb{R}$$ such that $$\forall n \in \mathbb{N}$$ we have that $$\frac{1}{|b_n|} \leq K$$. We can do this since the sequence is convergent and therefore bounded, so clearly the sequence $$\frac{1}{|b_n|}$$ will too be bounded.

Since $$\lim_{n \to \infty}a_n = a$$ then, $$\exists \: N_1 \in \mathbb{N}$$ such that $$\forall n \geq N_1$$ we have that $$|a_n - a| \leq \frac{\varepsilon}{2K}$$. Also since $$\lim_{n \to \infty}b_n = b$$ then, $$\exists \: N_2 \in \mathbb{N}$$ such that $$\forall n \geq N_2$$ we have that $$|b_n - b| \leq \frac{|b|}{|a|} \frac{\varepsilon}{2K}$$.

We have that \begin{align*}\left|\frac{a_n}{b_n} - \frac{a}{b}\right| &= \left|\frac{a_nb - ab_n}{b_nb}\right| \\ &= \left|\frac{a_nb - ba - ab_n + ba}{b_nb}\right| \\ &=\left|\frac{b(a_n - a)- a(b_n -b)}{b_nb}\right| \\ &= \left|\frac{(a_n-a)}{b_n} - \frac{a(b_n-b)}{b_nb}\right|\end{align*} Via the triangle inequality we have that \begin{align*}\left|\frac{a_n}{b_n} - \frac{a}{b}\right| =\left|\frac{(a_n-a)}{b_n} - \frac{a(b_n-b)}{b_nb}\right| \leq \left|\frac{a_n-a}{b_n}\right| + \left|\frac{a(b_n-b)}{b_nb}\right| = \frac{|a_n -a|}{|b_n|} + \frac{|a|}{|b|}\frac{|b_n-b|}{|b_n|}\end{align*}

Since $$|a_n - a| \leq \frac{\varepsilon}{2K}$$ then we have that $$\frac{|a_n -a|}{|b_n|} \leq \frac{\varepsilon}{|b_n|2K}$$ Also since $$|b_n - b| \leq \frac{|b|}{|a|}\frac{\varepsilon}{2K}$$ then $$\frac{|a|}{|b|}\frac{|b_n-b|}{|b_n|} \leq \frac{|b|}{|a|}\frac{\varepsilon}{2K}\cdot\left(\frac{|a|}{|b|}\frac{1}{|b_n|}\right) = \frac{\varepsilon}{2K|b_n|}$$

Since $$\frac{1}{|b_n|} \leq K$$ then $$\frac{\varepsilon}{2K|b_n|} \leq K\cdot\frac{\varepsilon}{2K} = \frac{\varepsilon}{2} \implies \frac{|a_n - a|}{|b_n|} \leq \frac{\varepsilon}{2}$$ \begin{align*} \frac{\varepsilon}{2K|b_n|} \leq \frac{\varepsilon}{2K}\cdot K = \frac{\varepsilon}{2} \implies\frac{|a|}{|b|}\frac{|b_n-b|}{|b_n|} \leq \frac{\varepsilon}{2}\end{align*}

So we can deduce that $$\frac{|a_n -a|}{|b_n|} + \frac{|a|}{|b|}\frac{|b_n-b|}{|b_n|} \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$

So then $$\left|\frac{a_n}{b_n} - \frac{a}{b}\right| \leq \varepsilon$$

This is true for $$N \in \mathbb{N}$$ when $$n \geq N$$ where $$N = max\{N_1, N_2\}$$, our choice of $$\varepsilon$$ was arbitrary so holds for all $$\varepsilon > 0$$. So indeed we have that $$\lim_{bn \to \infty}\frac{a_n}{b_n} = \frac{a}{b}$$.

• $b_n$ should not tend to $0$. Apr 5, 2021 at 19:40
• Thanks, I was aware of this, is the proof otherwise ok however? Apr 5, 2021 at 19:41
• @mosthigh It can’t be OK as nowhere you use the required hypothesis $b \neq 0$. Apr 5, 2021 at 19:43

$$\frac{1}{\vert b_n \vert} \le K$$ which is not correct.
The proposition is wrong by the way if $$b =0$$. You have to add the hypothesis $$b \neq 0$$ which implies as $$\{b_n\}$$ is supposed to be a convergent sequence that
$$\vert b_n \vert \ge \frac{\vert b \vert}{2}\gt 0$$ for $$n$$ large enough and then correct your proof accordingly.
• Your update is still not correct. You have in fact two constants. One, your original $K$ such that $\vert b_n \vert \le K$ valid for all $n$ because $\{b_n\}$ converges. And a second one $K^\prime$ such that $\frac{1}{\vert b_n \vert} \le K^\prime$ for $n$ large enough that exists because $\{b_n\}$ is supposed to converge to a non zero value. You need to explain that in your proof. Apr 6, 2021 at 5:57