# Thread: Another limit proof with divisibility of 2 sequences

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

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

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

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