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Thread: limit

  1. #1
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    limit

    I cant calculate this limit correctly...

    thank you very much for your help
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  2. #2
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    $\displaystyle \lim_{n\to\infty}(1-(\frac{2}{n+2})^{\frac{n+2}{n}}) = \lim_{n\to\infty}(1-(0)^1) = 1$
    Last edited by xxlvh; Nov 9th 2009 at 04:21 PM.
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  3. #3
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    thanks for the help but your answer is wrong...
    you answered for the limit of one multiplier...but you got n multipliers...so its not that simple...but thanks anyway
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  4. #4
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    nevemind I solved it...

    the limit is $\displaystyle e^-2$ by the way
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  5. #5
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    I realized that but upon inspection, every exponent should go to 0 as n approaches infinity, regardless of the number inside the brackets every multiplier will equal one. Try graphing a few terms, the limit is one. e^-2 is approximately 0.135 it's much too low.
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by UriK View Post
    nevemind I solved it...

    the limit is $\displaystyle e^{-2}$ by the way
    I'm inclined to agree. We have here

    $\displaystyle \lim_{n\to\infty}\prod_{k=1}^n\left(1-\frac{2}{k+2}\right)^{\frac{k+2}{n}}=\lim_{n\to\in fty}\prod_{k=1}^n\left[\left(1-\frac{1}{\frac{k+2}{2}}\right)^{\frac{k+2}{2}}\rig ht]^{\frac{2}{n}}$ $\displaystyle =\lim_{n\to\infty}\left[\prod_{k=1}^n\left(1-\frac{1}{\frac{k+2}{2}}\right)^{\frac{k+2}{2}}\rig ht]^{\frac{2}{n}}$

    Fact: $\displaystyle \left(1-\frac{1}{\alpha}\right)^{\alpha}$ tends to $\displaystyle e^{-1}$ as $\displaystyle \alpha$ gets large. (In this case $\displaystyle \alpha=\frac{k+2}{2}$.) Now the argument gets a bit informal (but hopefully stays correct), but as $\displaystyle n$ gets very large, the product becomes approximately $\displaystyle e^{-1}\cdot e^{-1}\cdot...e^{-1}=e^{-n}$. So we have:

    $\displaystyle \lim_{n\to\infty}\left[\prod_{k=1}^n\left(1-\frac{1}{\frac{k+2}{2}}\right)^{\frac{k+2}{2}}\rig ht]^{\frac{2}{n}}=\lim_{n\to\infty}(e^{-n})^{2/n}=\boxed{e^{-2}}$
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