# Thread: Another limit proof with divisibility of 2 sequences

1. ## Another limit proof with divisibility of 2 sequences

Let $(a_n)_{n \in \mathbb{N}}$ and $(b_n)_{n \in \mathbb{N}}$ be sequences, both greater then 0. Assume that $\frac{a_n}{b_n} \rightarrow X$ where $0. Show that $a_n \rightarrow \infty$ if and only if $b_n \rightarrow \infty$. [Hint: eventually $\frac{1}{2}X < \frac{a_n}{b_n} < \frac{3}{2}X$ ].

I was originally thinking of using the definition: $\left| \frac{a_n}{b_n} -X \right| <\epsilon$ but I realized that I was getting nowhere slowly. Looking at the hint I would think that $\left| \frac{a_n}{b_n} -X \right| <\frac{1}{2}$. At this point I don't know how to proceed.

2. $\begin{gathered}
\frac{X}
{2} > 0\; \Rightarrow \;\left( {\exists N} \right)\left[ {n \geqslant N\; \Rightarrow \;\left| {\frac{{a_n }}
{{b_n }} - X} \right| < \frac{X}
{2}} \right] \hfill \\
- \frac{X}
{2} < \frac{{a_n }}
{{b_n }} - X < \frac{X}
{2} \hfill \\
\end{gathered}$

Now just add $X$.