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Math Help - Limit of an Indeterminate Form Problem

  1. #1
    Junior Member
    Joined
    Jan 2010
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    28

    Limit of an Indeterminate Form Problem

    Hey I was just wondering if I did this problem right.

    \lim_{x \to \infty}(1 + \frac{1}{x})^x

    My Solution

    Let y = (1 + \frac{1}{x})^x, ln y = xln(1 + \frac{1}{x})<br />

    \lim_{x \to \infty} lny = \lim_{x \to \infty}\frac{ln(1 + \frac{1}{x})}{\frac{1}{x}} = \frac{0}{0}, Using L'Hopital's Rule.  = \lim_{x \to \infty}\frac{-\frac{1}{x^2 + x}}{-\frac{1}{x^2}} = \lim_{x \to \infty}\frac{x}{x + 1} = \frac{\infty}{\infty} = \lim_{x \to \infty}\frac{1}{1} = 1

    lny \to 1, y \to e

    Thanks in advance.
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  2. #2
    Senior Member
    Joined
    Dec 2008
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    Looks good to me.
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