I need some help finishing the idea for this proof
Suppose that there exist two sequences $(a_n),(b_n)$ such that the $\lim(a_n)=L_1$ and the $\lim(b_n)= L_2 \not = 0$. Prove that the $\lim(\frac{a_n}{b_n})= \frac{L_1}{L_2}$.
Solution: Observe that the as ($a_n)$ converges then $\forall \epsilon > 0, \exists N_1 \in \mathbb{N}$ such that $\forall n \geq N_1$
$$\left\vert a_n – L_1\right\vert< \epsilon$$
Similarly, as $(b_n)$ converges, then $\forall \epsilon > 0, \exists N_2 \in \mathbb{N}$ such that $\ n \geq N_2$
$$\left\vert b_n – L_2\right\vert < \epsilon$$
Then if the $(\frac{a_n}{b_n})$ converges then, $\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that $\forall n \geq N$
$$\left\lvert\frac{a_n}{b_n}-\frac{L_1}{L_2}\right\rvert < \epsilon$$
$$\left\vert\frac{a_n}{b_n}-\frac{L_1}{L_2}\right\vert = \left\vert\frac{a_n \cdot L_2 – b_n \cdot L_1}{b_n \cdot L_2}\right\vert < \epsilon$$
Add\Subtract $L_1 \cdot L_2$ to the LHS to express
$$\left\vert\frac{a_n \cdot L_2 – L_1 \cdot L_2 + L_1 \cdot L_2 – b_n \cdot L_1}{b_n \cdot L_2}\right\vert = \left\vert\frac{L_2(a_n -L_1) +L_1 (L_2- b_n)} {b_n \cdot L_2}\right\vert$$
$$=\left\vert\frac{1}{b_n} \cdot (a_n -L_1) – \frac{1}{b_n} \cdot \frac{L_1}{L_2} \cdot (b_n -L_2)\right\vert < \epsilon $$
As the LHS of the inequality is a measure of distance between terms, we use the Triangle Inequality to express
$$=\left\vert\frac{1}{b_n} \cdot (a_n -L_1) – \frac{1}{b_n} \cdot \frac{L_1}{L_2} \cdot (b_n -L_2)\right\vert \leq \left\vert\frac{1}{b_n} \cdot (a_n -L_1)\right\vert + \left\vert\frac{1}{b_n} \cdot \frac{L_1}{L_2} \cdot (b_n -L_2)\right\vert $$
$$=\left\vert\frac{1}{b_n}\right\vert \cdot \left\vert(a_n -L_1)\right\vert + \left\vert\frac{1}{b_n}\right\vert\cdot \left\vert\frac{L_1}{L_2}\right\vert \cdot \left\vert(b_n -L_2)\right\vert < \epsilon$$
Now we know from the initial conditions that there exist a $n \geq N_1, N_2$ respectively such that the expressions hold $\forall \epsilon >0$.
I understand how to build the argument using $\frac{\epsilon}{2}$ and why we can use this.
Im stuck on $\left\vert\frac{1}{b_n}\right\vert$, we know that $b_n$ converges, which implies there must exist a bound such that
$$M \in \mathbb{R}, \left\vert b_n\right\vert \leq M$$
How does this bound work with $\left\vert\frac{1}{b_n}\right\vert$? what do i need to do to find the bound i need.
Best Answer
Because $\lim b_n$ exists, we can choose $\epsilon=|\lim b_n|/2$(here we are using the non zero condition), by definition there exists $N_0$ such that for all $n>N_0$ we have $|b_n-\lim b_n|<|\lim b_n|/2$. Can you bound $\frac{1}{|b_n|}$ from here?