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Thread: Properly Divergent Limits

  1. #1
    Junior Member
    Feb 2008

    Properly Divergent Limits

    $\displaystyle \lim\nolimits_{n\to\infty}$$\displaystyle (x_n)= \infty$ and let $\displaystyle (y_n)$ be such that $\displaystyle lim(x_ny_n)$ belongs to $\displaystyle R$. Show that $\displaystyle (y_n)$ converges to 0.

    So I know that there is some K($\displaystyle \epsilon$) such that both for all n>K$\displaystyle \epsilon$ $\displaystyle x_n>\alpha$ and

    $\displaystyle |x_ny_n - x|< \epsilon$

    I know I can manipulate this with division, but anyway I try this, I can't get the result
    $\displaystyle |y_n - 0|< \epsilon$

    my simplification is

    |$\displaystyle y_n - 0$|$\displaystyle < \frac{\epsilon +x}{x_n}$

    Could someone please help?
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  2. #2
    MHF Contributor

    Aug 2006
    Suppose that $\displaystyle \varepsilon > 0\,\& \,\left( {x_n y_n } \right) \to x$
    $\displaystyle \left( {\exists N_1 } \right)\left[ {n \geqslant N_1 \Rightarrow \left| {x_n y_n - x} \right| < 1 \Rightarrow \left| {x_n y_n } \right| < 1 + \left| x \right|} \right]$.
    Now $\displaystyle \left( {\exists K>0} \right)\left[ {\frac{{1 + \left| x \right|}}
    {K} < \varepsilon } \right]$.
    Because $\displaystyle \left( {x_n } \right) \to \infty \Rightarrow \left( {\exists N_2 } \right)\left[ {n \geqslant N_2 \Rightarrow x_n > K} \right]$.
    $\displaystyle n \geqslant N_1 + N_2 \Rightarrow \quad \left| {y_n } \right| < \frac{{1 + \left| x \right|}}{{\left| {x_n } \right|}} < \frac{{1 + \left| x \right|}}{K} < \varepsilon$
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