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

  • Thread starter mehgunn
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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?
 

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  • #2
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How exactly did you apply l'Hopital's theorem? Because it does not give you [itex](\log x)^2[/itex].
 
  • #3
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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.
 
  • #4
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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.
 
  • #5
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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?

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.
 
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  • #6
lavinia
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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?
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.
 
  • #7
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[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]
 
  • #8
Curious3141
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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.
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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