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Thread: Help solving this limit

  1. March 10th 2010, 09:19 AM #1
    paupsers
    paupsers is offline
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    Help solving this limit

    lim_{k-->\infty} (\frac{k-2}{k})^k

    When you just "plug in," you get 1^\infty which is indeterminate. So then I rewrote it as:

    exp[kln(\frac{k-2}{k})=exp[kln(k-2)]exp[-klnk] and this is where I'm stuck. Any ideas? Is there a simpler way to do this?
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  2. March 10th 2010, 09:48 AM #2
    qmech
    qmech is offline
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    lim_{k->\infty} (\frac{k-2}{k})^k

    = lim_{k->\infty} (1 + \frac{-2}{k})^k

    = e^{-2}


    Remember:
    = lim_{k->\infty} (1 + \frac{x}{k})^k = e^x
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