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Thread: Another limit proof with divisibility of 2 sequences

  1. #1
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    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.
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  2. #2
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    $\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 $\displaystyle X$.
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