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Math Help - Trouble on my mind

  1. #1
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    Trouble on my mind

    I didn't solve it pls help me

    Problem:

    Limit x->0+ [(x+e^x)^(1/x)]

    Tanks for your interest
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  2. #2
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    Re: Trouble on my mind

    find the limit of the natural log of the expression and take the exponential of that limit.

    $\ln\left((x+e^x)^{1/x}\right) = \dfrac 1 x \ln(x+e^x)$

    Now apply L'Hopital's rule

    $f(x)=\ln(x+e^x)$

    $g(x)=x$

    $f'(x)=\dfrac{1+e^x}{x+e^x}$

    $g'(x)=1$

    $\displaystyle{\lim_{x \to 0}}\dfrac{f'(x)}{g'(x)}=\dfrac{1+1}{0+1}=2$

    Now take the exponential of this to obtain the limit of the original expression.

    $\displaystyle{\lim_{x\to 0}}\left((x+e^x)^{1/x}\right) =e^2$
    Last edited by romsek; April 12th 2014 at 12:38 AM.
    Thanks from JeffM
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  3. #3
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    Re: Trouble on my mind

    Tanks but we don't know "Taylor series". Have you another solution?
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  4. #4
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    Re: Trouble on my mind

    I editted the solution. I should have used L'Hopital's rule to begin with.
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  5. #5
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    Re: Trouble on my mind

    This is it! Tanks a lot
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