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Math Help - L'Hopital Limit

  1. #1
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    L'Hopital Limit

    \lim_{x\to\infty}x{[(1+\frac{1}{x})^x -e]} I know its indeterminate form \infty*0 and so i rewrite it as
    \lim_{x\to\infty}\frac{(1+\frac{1}{x})^x -e}{\frac{1}{x}}. Now it's \frac{0}{0} indeterminate form. Then I get stuck.
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  2. #2
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    Quote Originally Posted by xxsteelxx View Post
    \lim_{x\to\infty}x{[(1+\frac{1}{x})^x -e]} I know its indeterminate form \infty*0 and so i rewrite it as
    \lim_{x\to\infty}\frac{(1+\frac{1}{x})^x -e}{\frac{1}{x}}. Now it's \frac{0}{0} indeterminate form. Then I get stuck.

    This is really a tough limit: first, and in order to derivate more easily, put

    \displaystyle {\left(1 + \frac{1}{x}\right)^x= e^{x\ln \left(1 + \frac{1}{x}\right)}}

    so that \left(\left(1 + \frac{1}{x}\right)^x\right)' =\left(1 + \frac{1}{x}\right)^x \left[\ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right]

    Applying now L'Hospital to your limit we get -x^2\left(1 + \frac{1}{x}\right)^x \left[\ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right].

    Let apart the -\left(1 +\frac{1}{x}\right)^x term, whose limit is -e , and focus on the rest:

    \frac{\ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}}{\frac{1}{x^2}}

    Apply L'Hospital one again here and get \frac{-\frac{1}{x(x+1)}+\frac{1}{(x+1)^2}}{-\frac{2}{x^3}} =\frac{1}{2}\left[\frac{x^2}{x+1}-\frac{x^3}{(x+1)^2}\right] =\frac{1}{2}\frac{x^2}{(x+1)^2} \xrightarrow [x \to \infty] {} \frac{1}{2}

    Now don't forget the term you left out before and the limit finally is (**Pant, pant!!**) : -\frac{e}{2}

    Tonio
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  3. #3
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    Thank you very much!
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