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L'hospital's rule

  1. Nov 20, 2007 #1
    1. The problem statement, all variables and given/known data
    lim (1-10x)^(1/x)
    x->0
    evaluate the limit

    2. Relevant equations
    L'hostpital's rule


    3. The attempt at a solution

    take derivative:
    lim (-10+100x)/x
    x->0

    can't divide by zero so take the derivative again but x goes away:
    lim 100
    x->0

    is 100 the limit? is there a limit? now that i'm looking at it again, i dont think i have the derivative right...
     
  2. jcsd
  3. Nov 20, 2007 #2
    Take natural logs on both sides. You'll get,

    ln(L) = lim(x->0) {1/x*ln(1-10x)} which is of the form 0/0. Apply l'Hospital's rule now. When done, convert the ln(L) = m {where m is the value of limit you got} again into exponential form, i.e. L=e^m.

    Regards,
    Sleek.
     
  4. Nov 20, 2007 #3

    HallsofIvy

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    WHAT did you take the derivative of? I don't recognize that as having anything to do with your original limit!

    Since your original form is NOT f(x)/g(x), the first thing I would do it take the logarithm:
    If Y= [tex](1-10x)^{1/x}[/tex] then ln(Y)= ln(1-10x)/x. Now apply L'Hopital's rule to that.
     
  5. Nov 21, 2007 #4
    ok...hows this look?

    ln(L) = lim(x->0) { ln(1-10x)/x } = 0/0 so..

    ln(L) = lim(x->0) { (-10)/(1-10x) }

    ln(L) = -10

    L = e^-10
     
  6. Nov 21, 2007 #5

    HallsofIvy

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    Much better.

    (Sorry about posting the same thing so many times. I got a bit carried away, didn't I?)
     
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