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Can i use bernoulis inequality like this?

  1. Oct 23, 2005 #1
    can I use bernoulis inequality like this for finding this limit?

    [tex]\lim_{n\rightarrrow\infty}(1+\frac{1}{2n}) \geq \lim (1+2n(\frac{1}{2n})) \geq 2[/tex]
     
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  3. Oct 23, 2005 #2

    HallsofIvy

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    Well,first of all, you aren't using Bernoulli's inequality.
    Bernoulli's inequality says that
    [tex](1+x)^r\ge 1+rx[/tex]
    as long as x and r are both larger than -1.

    In this case, r= 1 so Bernoulli's inequality doesn't say anything.
    In any case, isn't it obvious that, as [itex]x->\infty[/itex],
    [tex]\frac{1}{2n}-> 0[/tex]?
    And so
    [frac]1+ \frac{1}{2n}\rightarrow 1[/tex].

    Or did you mean
    [tex]lim_{n\rightarrow\infty}\left(1+\frac{2n}\right)^{2n}[/tex]?
    Yes, Bernoulli's inequality applies to this and shows that the limit is greater than 2. In fact, it should be clear that the limit is e which is certainly larger than 2!
     
  4. Oct 23, 2005 #3
    your right i did have a typo i meant
    [tex]\lim_{n\rightarrrow\infty}(1+\frac{1}{2n})^{2n} \geq \lim (1+2n(\frac{1}{2n})) \geq 2[/tex]
     
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