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Math Help - Limit of an Indeterminate Form

  1. #1
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    Limit of an Indeterminate Form

    Hey guys, I have a problem that I'm having trouble with. Any help would be appreciated.

    \lim{x}\rightarrow\infty (1 + \frac{1}{x})^x = Indeterminate Form 1^\infty

    Let y = (1 + \frac{1}{x})^x, lny = ln(1 + \frac{1}{x})^x = xln(1 + \frac{1}{x})

    \lim{x}\rightarrow\infty lny = \lim{x}\rightarrow\infty xln(1 + \frac{1}{x}) = \lim{x}\rightarrow\infty \frac{ln(1 + \frac{1}{x})}{\frac{1}{x}} = Indeterminate Form \frac{0}{0} = (Using L'Hopital's Rule) \lim{x}\rightarrow\infty \frac{\frac{1}{1 + \frac{1}{x}}}{\frac{-1}{x^2}}

    I'm stuck here. The limit of the top is 1 and the limit of the bottom is 0. I don't know if I did something wrong or if I just don't know what to do next. Thanks in advance.
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  2. #2
    Super Member Aryth's Avatar
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    I'm going to start here:

    \lim_{x \to \infty} \frac{\ln{\left(1 + \frac{1}{x}\right)}}{\frac{1}{x}}

    Using L'Hopitals rule, you must use the Quotient Rule, which will yield:

    \frac{\left(\ln{\left(1 + \frac{1}{x}\right)}\right)'\frac{1}{x} - \ln{\left(1 + \frac{1}{x}\right)}\left(\frac{1}{x}\right)'}{\lef  t(\frac{1}{x}\right)^2}

    The result from this will yield a different form than the one you have provided.

    That should lead you to the answer.

    Good Luck.
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