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Math Help - Properly Divergent Limits

  1. #1
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    Properly Divergent Limits

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

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

    |x_ny_n - x|< \epsilon

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

    my simplification is

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

    Could someone please help?
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  2. #2
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    Suppose that \varepsilon  > 0\,\& \,\left( {x_n y_n } \right) \to x
    \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 \left( {\exists K>0} \right)\left[ {\frac{{1 + \left| x \right|}}<br />
{K} < \varepsilon } \right].
    Because \left( {x_n } \right) \to \infty  \Rightarrow \left( {\exists N_2 } \right)\left[ {n \geqslant N_2  \Rightarrow x_n  > K} \right].
    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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