Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having a hard time understanding why this limits equals to 0.

By simply using logarithm identity I get that this limit equals

$$ \lim_{x\to 0} \ e^{x\ln x^{x}}=\lim_{x\to 0}\ e^{x^{2}\ln x}$$ and by L'Hopital we get that the limit of the exponent is 0 and because $f(a)=e^{a}$ is continuous I get that the final limit should be 1.

I'd love your help with understanding what I did wrong.

Thank you.

share|improve this question
3  
Your use of what you call logarithm identity is wrong, you mixed up $x$ and $x^x$. –  Did Jun 21 '11 at 15:36
    
There is a general convention that $x^{x^x}$ means $x^{(x^x)}$. But not everyone is always aware of the convention. I think it is better to put in the parentheses, even if they look ugly. –  André Nicolas Jun 21 '11 at 16:17
    
Yeah, I should have been more careful with this, I guess it was more comfortable :-) –  user6163 Jun 21 '11 at 16:25
    
@user6312 : The way I remember this convention is that $\left(x^y\right)^z = x^{(yz)}$. So since there is not much use in having two different notations for the same thing, we set $x^{y^z} = x^{\left(y^z\right)}$. –  Joel Cohen Jun 21 '11 at 20:12
    
@Joel Cohen: A good way of remembering. My way is "its the really really fast one." –  André Nicolas Jun 21 '11 at 20:23

2 Answers 2

up vote 9 down vote accepted

You considered $(x^x)^x$ instead of $x^{(x^x)}$.

share|improve this answer
    
Thank you; –  user6163 Jun 21 '11 at 15:44
    
Thanks (and your proof is fine for $\lim _{x \to 0^ + } (x^x )^x = 1$). –  Shai Covo Jun 21 '11 at 15:59
1  
Note that, in any case, the limit should be one-sided, since $x^x$ is not defined on the negative real axis. –  Shai Covo Jun 21 '11 at 16:38

The mistake you made is parenthesis in the exponent. In any case, here is another way to solve this problem:

Since $\lim_{x\rightarrow 0^+} x^x =1$, there is an interval $(0,\delta)$, $0<\delta<1$ such that $\frac{1}{2}<x^x<\frac{3}{2}$. Then on this interval, $$x^{\frac{3}{2}}\leq x^{x^x} \leq x^{\frac{1}{2}}$$ so we see the limit is $0$ by the Squeeze Theorem.

Hope that helps,

share|improve this answer
    
very nice, thanks. –  user6163 Jun 21 '11 at 15:52
1  
minor point, but the limit should be a one-sided limit, of course. –  Shai Covo Jun 21 '11 at 16:25
    
@Shai: Thanks, I fixed it now. –  Eric Naslund Jun 21 '11 at 16:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.