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Limit with ln in the exponent

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data
    lim as x approaches 0 of (x + 1)ln x

    2. Relevant equations



    3. The attempt at a solution
    I tried using ln y = (ln x)[ln (x + 1)]
    then:
    ln (x + 1)
    ----------
    1
    ---
    ln x
    to make it eligible for L'Hopital's Rule. Then differentiating the numerator and the denominator, I got:
    1
    ---
    x + 1
    -------
    -1
    ---------
    x(ln x)2
    Then I brought the denominator of the denominator up:
    -x(ln x)2
    ----------
    x + 1

    The answer is supposed to be 1. Therefore my differentiating should have evaluated to 0. Doesn't my answer give -infinity? Please help. I suck at ln and e. I must have missed something.
     
    Last edited: Apr 29, 2009
  2. jcsd
  3. Apr 29, 2009 #2

    Dick

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    -x(ln x)^2/(x+1) is still an indeterminant form as x->0, ln(x)->-infinity. You'll have to apply l'Hopital again to figure out the limit of the numerator.
     
  4. Apr 29, 2009 #3
    Thank you for replying.

    I did that, and I got: - 2 ln x - (ln x)2, which is inf - inf. I'm rather confused how to convert that to numerator/denominator form again, though.
     
    Last edited: Apr 29, 2009
  5. Apr 29, 2009 #4

    Dick

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    You want the limit of -x*ln(x)^2. First write it as -ln(x)^2/(1/x). I'm not sure how you got what you got.
     
  6. Apr 29, 2009 #5
    Ah. My mistake. You said "numerator". I hadn't realized that once you'd found the limit of either numerator or denominator you could go to work on each separately (I thought if you used L'Hopital you always had to differentiate both haha).

    Anyway, after a couple more differentiations of the numerator I eventually got -2x which would indeed make the whole thing zero. Thanks. :!!)
     
  7. Apr 29, 2009 #6

    Dick

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    Right. Once you know the limit of a part of an expression you can set it aside and just work with the rest using l'Hopital. It simplifies the expressions.
     
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