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Simple limit

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data

    lim x->0 (n/log(n))[n1/n-1]

    3. The attempt at a solution

    I've just been trying to move things around, and use L'Hospitals when appropriate. I haven't been able to see the trick.

    Got any hints?
  2. jcsd
  3. Mar 9, 2009 #2
    Try the algebra of limits
  4. Mar 9, 2009 #3
    Do you mean lim n->infinity?
  5. Mar 10, 2009 #4
    woops, ya as n goes to infinity.

    Ive been using L'hospitals on the numerator after multiplying the n through and rewriting in exponential form.

    d/dx:[x1/x+1-x] = d/dx:[e(1/x+1)ln(x)-x]= e(1/x+1)ln(x)( -ln(x)/x2+(x+1)/x2) -1
    and the bottom goes to 1/x

    so the new limit after applying L'Hospital is
    e(1/x+1)ln(x)((x+1)-ln(x)/x) -x

    which leaves me with a big question mark.
  6. Mar 10, 2009 #5
    also tried looking at it like this.


    using L'H on the top and bottom give:

    x1/x((1-ln(x)/x2)/(1-log(x))... I want to do something with that log
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