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Thread: [SOLVED] L'Hopital Rule II

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Question [SOLVED] L'Hopital Rule II

    I was working on another problem, and I get $\displaystyle \frac {1 + \frac{1}{x}}{1}$ and the limit as it approaches x is 0, but apparently the answer here is 2

    how? what about $\displaystyle \frac {1}{x}$
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    I was working on another problem, and I get $\displaystyle \frac {1 + \frac{1}{x}}{1}$ and the limit as it approaches x is 0, but apparently the answer here is 2

    how? what about $\displaystyle \frac {1}{x}$
    what is the original problem?
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  3. #3
    Member >_<SHY_GUY>_<'s Avatar
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    it was $\displaystyle (e^x + x)^\frac{1}{x}$
    as x approaches 0
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  4. #4
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    it was $\displaystyle (e^x + x)^\frac{1}{x}$
    as x approaches 0
    let $\displaystyle y = (e^x + x)^{\frac{1}{x}}$

    $\displaystyle \ln{y} = \frac{\ln(e^x + x)}{x}$

    $\displaystyle \lim_{x \to 0} \frac{\ln(e^x + x)}{x} =$

    $\displaystyle \lim_{x \to 0} \frac{e^x + 1}{e^x + x} = 2$

    since $\displaystyle \lim_{x \to 0} \ln{y} = 2$ , $\displaystyle \lim_{x \to 0} y = e^2$
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  5. #5
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    here's another approach by using elementary limits:

    for $\displaystyle x\ne0$ let $\displaystyle \left( e^{x}+x \right)^{\frac{1}{x}}=\exp \left( \frac{\ln \left( 1-\left( 1-\left( e^{x}+x \right) \right) \right)}{1-\left( e^{x}+x \right)}\cdot \frac{1-\left( e^{x}+x \right)}{x} \right),$ thus as $\displaystyle x\longrightarrow0$ this tends to $\displaystyle \exp \left( -1\times \left( -1-1 \right) \right)=e^{2}.$
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