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Math Help - Calculation of a limit using L'Hopital

  1. #1
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    Calculation of a limit using L'Hopital

    Hello


    I'm having trouble calculating the right hand limit of sqrt(x) * ln(x) at 0

    I see that it's an indeterminate form of the type 0 * -infinity
    So I tried turning into the limit of sqrt(x) / (1/lnx) so that it would be a case of 0/0
    and applied L'Hopitals rule, but I got that it's equal to
    1/2 * limit at 0+ of sqrt(x) * (lnx)^2

    In other words back where I started...what am I doing wrong?
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  2. #2
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    Quote Originally Posted by moses View Post
    Hello


    I'm having trouble calculating the right hand limit of sqrt(x) * ln(x) at 0

    I see that it's an indeterminate form of the type 0 * -infinity
    So I tried turning into the limit of sqrt(x) / (1/lnx) so that it would be a case of 0/0
    and applied L'Hopitals rule, but I got that it's equal to
    1/2 * limit at 0+ of sqrt(x) * (lnx)^2

    In other words back where I started...what am I doing wrong?
    Try doing it the other way.
    \sqrt{x}ln(x) = \frac{ln(x)}{\frac{1}{\sqrt{x}}}

    -Dan
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  3. #3
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    Oh okay that works, thanks
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