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Strange l'Hospital's Rule with ln

  1. Jul 10, 2010 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    Since lim(x→∞)(lnx )^2=∞ and lim(x→∞)x=∞, the function is indeterminate, ∞/∞.

    lim(x→∞)(lnx)^2/x=lim(x→∞) d/dx (lnx )^2/(d/dx x)=lim(x→∞)(2 lnx (1/x))/1

    An I am stuck...
  2. jcsd
  3. Jul 10, 2010 #2
    I think L'Hostpital's rule can be used very easily here. - just apply it again.

    Alternatively, it's not hard to see the answer from the following expansion of logarithm.

    [tex]{\rm ln}(x)=2\sum_{i=1}^\infty {{1}\over{2i-1}}\Bigl({{x-1}\over{x+1}}\Bigr)^{2i-1}[/tex]

    With experience, you will recognize the answer to similar limits by inspection. Not many functions increase monotonically as slowly as logarithm. Granted, that's not a reliable, nor mathematically correct way to get the answer, but it helps to know the answer before you solve the problem.
  4. Jul 10, 2010 #3
    This might be a higher-level answer. My answer should be looking much easier cause we don't work with i's and sums.
  5. Jul 10, 2010 #4
    OK, don't worry about that. Just take your first answer from L'Hospital's rule and put it in a simple numerator over denominator form. Then apply the rule again.
  6. Jul 10, 2010 #5
    (2 lnx (1/x))/1 = 2(ln x)/x

    Can you finish it from there?
  7. Jul 10, 2010 #6
    Okay, so here it is:
    lim(x→∞) (2 lnx)/x=
    lim(x→∞)(d/dx lnx)/(d/dx x)=
    lim(x→∞)(2 (1/x))/1=

    Should I take a derivative again, or it's enough and 2/infinity equals zero?
  8. Jul 10, 2010 #7
    You have it, the answer is 0. You can't use l'Hopital's rule again anyway since 2/x isn't indeterminate like the other expressions you had before.
  9. Jul 10, 2010 #8
    I see it now, thank you very much.
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