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Math Help - Limit Problem - # 2

  1. #1
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    Limit Problem - # 2

    \lim x \rightarrow \infty

    x^{\dfrac{4}{x}}

    \infty^{\dfrac{4}{\infty}}

    \infty^{0}

    Quotient Rule or not - I think not.

    With Quotient Rule

    \ln[x^{\dfrac{4}{x}}]


    \dfrac{4\ln x}{x}

    \dfrac{(x)(\dfrac{4}{x} - \ln x (1)}{x^{2}}

    \dfrac{4 (\infty) - 4 \ln (\infty)}{\infty^{2}}

    Without Quotient

    \ln[x^{\dfrac{4}{x}}]


    \dfrac{4 \ln x}{x}

    \dfrac{\dfrac{d}{dx} 4 \ln x}{\dfrac{d}{dx} x}

    \dfrac{4 \ln (\infty)}{(\infty)}

    \dfrac{(\infty)}{(\infty)} = Indeterminate What now?
    Last edited by Jason76; October 30th 2013 at 09:16 PM.
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  2. #2
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    Re: Limit Problem - # 2

    Limit Problem - # 2-31-oct-13.pngClick image for larger version. 

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    Last edited by ibdutt; October 30th 2013 at 10:10 PM.
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  3. #3
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    Re: Limit Problem - # 2

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

    x^{\dfrac{4}{x}}

    \infty^{\dfrac{4}{\infty}}

    \infty^{0}

    Quotient Rule or not - I think not.

    With Quotient Rule

    \ln[x^{\dfrac{4}{x}}]


    \dfrac{4\ln x}{x}

    \dfrac{(x)(\dfrac{4}{x} - \ln x (1)}{x^{2}}

    \dfrac{4 (\infty) - 4 \ln (\infty)}{\infty^{2}}

    Without Quotient

    \ln[x^{\dfrac{4}{x}}]


    \dfrac{4 \ln x}{x}

    \dfrac{\dfrac{d}{dx} 4 \ln x}{\dfrac{d}{dx} x}

    \dfrac{4 \ln (\infty)}{(\infty)}

    \dfrac{(\infty)}{(\infty)} = Indeterminate What now?
    \displaystyle \begin{align*} \lim_{x \to \infty} x^{\frac{4}{x}} &= \lim_{x \to \infty} e^{\ln{ \left[ x^{\frac{4}{x}} \right] }} \\ &= \lim_{x \to \infty} e^{ \frac{4\ln{(x)}}{x} } \\ &= e^{ \lim_{x \to \infty} \frac{4\ln{(x)}}{x} } \end{align*}

    This is now in a form where L'Hospital's Rule can be applied.
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