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What is the limit of x^lnx as x approaches 0+?

  1. Jan 22, 2012 #1
    It seems simple enough, but I keep getting stuck.

    When originally taking the limit I end up with 0^∞, which is an indeterminate form.

    I have tried to simplify the problem by taking the natural log of it, which leaves me with lnx*lnx. After applying L'Hospitals Theorem I still end up with lnx*lnx.

    Can anyone help me figure out where I am going wrong?
  2. jcsd
  3. Jan 22, 2012 #2
    How exactly did you apply l'Hopital's theorem? Because it does not give you [itex](\log x)^2[/itex].
  4. Jan 22, 2012 #3
    Well, I know that e^ln(x^lnx) is the same as x^lnx, but I put aside the e for the end of the equation.
    ln(x^lnx)= lnx*lnx.

    To apply L'Hospitals Rule I have to have it in quotient form, so I rearranged my equation to lnx/ (1/lnx).

    After taking the derivative of both I ended up at -(lnx)^2.
  5. Jan 22, 2012 #4
    First off lnx goes to -∞ as x goes to 0+, not ∞. With this in mind remember that we are only interested in values as x gets extremely close to zero from above, we don't care what happens at x=0. Thus try evaluating x^lnx for .1 then .01 then .001 and observe its behavior, finally finish off with a δ-M proof.
  6. Jan 22, 2012 #5

    x^log(x) = e^(log(x) * log(x)) = e^((log(x))^2)

    As x -> 0, log(x) goes to -infinity so (log(x))^2 goes to infinity, and so does e^that.

    Then I put it in a graphing calculator and verified the result visually.

    You were on the right track with log(x) * log(x). As x->0, log(x) goes to -inf so log(x) * log(x) goes to +inf. Exponentiating to get back to where you started (since you originally took the log) you get the same result I did. And you can justify logging and exp-ing since continuous functions preserve limits.

    But rather than initially take the log, it's better to start by expressing any exponential expression as e^something using a^b = e^(b * log(a)). That's the first thing to do whenever you see an exponential expression where the base isn't already e.

    BTW I use "log" to mean natural log since that's the usual convention in math.
    Last edited: Jan 22, 2012
  7. Jan 22, 2012 #6


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    the log diverges continuously to negative infinity as x approaches zero. So x^ln(x) is the reciprocal of a very small number raised to a huge power.
  8. Jan 22, 2012 #7
    [itex]\displaystyle\lim_{x\to 0^+}x^{ln(x)}[/itex]

    [itex]Let \ \ x \ = \ e^y[/itex]

    [itex] Then \ \ ln(x) \ = \ y[/itex]

    [itex]As \ \ x \rightarrow 0^+, \ \ e^y \ \rightarrow \ 0^+[/itex]

    [itex]Then \ \ y \ \rightarrow \ -\infty[/itex]

    The limit can be rewritten as:

    [itex]\displaystyle\lim_{y\to -\infty}e^{y^2} \ = [/itex]

  9. Jan 23, 2012 #8


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    Since noone has yet commented on this, let me correct you. L' Hopital's Rule can only be applied when the limit is of the indeterminate form 0/0 or ±∞/∞.

    In this case ln x/(1/ln x) is not of the correct form (it's of the form -∞/0-), and LHR does not apply.

    You should just take the limit of ln x, which is -∞ and square it to get ∞, as SteveL27 has pointed out.
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