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Math Help - Limit Problem

  1. #1
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    Limit Problem

    \lim x \rightarrow \infty

    (e^{x} + x)^{\dfrac{9}{x}}

      (e^{\infty} + \infty)^{\dfrac{9}{\infty}}

    (\infty + \infty)^{0}

    (\infty)^{0}
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  2. #2
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    Re: Limit Problem

    Given \lim_{x\rightarrow \infty}\left(e^x+x\right)^{\frac{9}{x}}

    Now when x\rightarrow \infty, Then e^x>>>>>>>>>x

    So \lim_{x\rightarrow \infty}\left(e^x+x\right)^{\frac{9}{x}}\approx \lim_{x\rightarrow \infty}\left(e^x\right)^{\frac{9}{x}} = e^9
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  3. #3
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    Re: Limit Problem

    Quote Originally Posted by Jason76 View Post
    \lim x \rightarrow \infty

    (e^{x} + x)^{\dfrac{9}{x}}

      (e^{\infty} + \infty)^{\dfrac{9}{\infty}}

    (\infty + \infty)^{0}

    (\infty)^{0}
    For a more sophisticated approach

    \displaystyle \begin{align*} \left( e^x + x \right) ^{\frac{9}{x}} &= e^{\ln{\left[ \left( e^x + x \right) ^{\frac{9}{x}} \right] }} \\ &= e^{\frac{9\ln{ \left( e^x + x \right) }}{x}} \end{align*}

    and so

    \displaystyle \begin{align*} \lim_{x \to \infty} \left[ \left( e^x + x \right) ^{\frac{9}{x}} \right] &= \lim_{x \to \infty} e^{ \frac{9\ln{ \left( e^x + x \right) }}{x} } \\ &= e^{ \lim_{x \to \infty} \frac{9\ln{ \left( e^x + x \right) }}{x} } \\ &= e^{ \lim_{x \to \infty} \frac{\frac{9 \left( e^x + 1 \right) }{e^x + x}}{1} } \textrm{ by L'Hospital's Rule} \\ &= e^{ \lim_{ x \to \infty} \frac{9 \left( e^x + 1 \right) }{e^x + x} } \\ &= e^{\lim_{x \to \infty} \frac{9e^x}{e^x + 1} } \textrm{ by L'Hospital's Rule again} \\ &= e^{\lim_{x \to \infty} \frac{9e^x}{e^x}  } \textrm{ by L'Hospital's Rule again} \\ &= e^{\lim_{x \to \infty} 9 } \\ &= e^9 \end{align*}
    Thanks from topsquark
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