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L'Hopital's rule of x*ln(x)-x

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    What is the value of xln(x)-x when x=0?


    2. Relevant equations
    I'm assuming you do L'Hopital's


    3. The attempt at a solution
    I'm assuming you factor out the x, leaving:

    x(ln(x)-1)

    but that's still not in the form of [tex]\frac{\infty}{\infty}[/tex] or [tex]\frac{0}{0}[/tex]

    Would you do:

    [tex]lim_{x\rightarrow\infty}[/tex][tex]\frac{(ln(x)-1)}{x^{(-1)}}[/tex]

    =[tex]lim_{x\rightarrow\infty}[/tex][tex]\frac{(1/x)}{1}[/tex]

    =0

    ??
     
  2. jcsd
  3. Mar 17, 2009 #2

    gabbagabbahey

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    [tex]\lim_{x\to 0} x\ln(x)-x=(\lim_{x\to 0} x\ln(x))-(\lim_{x\to 0} x)[/tex] :wink:
     
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