Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    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]


    [itex]\infty[/itex]
     
  9. Jan 23, 2012 #8

    Curious3141

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What is the limit of x^lnx as x approaches 0+?
Loading...