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I've encountered the following problem: Evaluate $$\lim_{x\to 0^+}\cfrac{x^{x^x}}{x}.$$

This is readily a "$\frac{0}{0}$" form, so I used L'Hopital's rule, but it got seriously messy, and fast. Can anyone recommend an alternative approach?

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Just for you to be safe, the limit is 1. –  Pedro Tamaroff Jun 5 '12 at 2:48
    
Good to know! Thank you, @Peter. –  Cameron Buie Jun 5 '12 at 2:50
    
Peter suggested, and I agree, that you perhaps consider the limit of $\log\left({x^{x^x}\over x}\right)$ instead, or perhaps the log of that. –  MJD Jun 5 '12 at 2:51
    
@MarkDominus: Yeah, I gave that a go, too, and hit $$\cfrac{x^x-1}{1/\log x}$$ with L'Hopital's rule. Again quite messy, and I couldn't see a natural way to rearrange that so that iterations would make it simplifiable. Perhaps I'm just missing something obvious. –  Cameron Buie Jun 5 '12 at 2:55
    
@CameronBuie It is desirable not to use \dfrac, \cfrac, \displaystyle in the titles since it breaks the layout of the list of questions. I have changed it to \frac –  user17762 Jun 5 '12 at 3:54

5 Answers 5

up vote 19 down vote accepted

You can avoid series by taking advantage of the known limit $\lim_{x\to 0^+}x^x=1$. Let $f(x)=\ln\left(\dfrac{x^{x^x}}x\right)=(x^x-1)\ln x$. Then

$$\begin{align*} \lim_{x\to 0^+}f(x)&=\lim_{x\to 0^+}(x^x-1)\ln x\\\\ &=\lim_{x\to 0^+}\frac{\ln x}{\frac1{x^x-1}}\\\\ &=\lim_{x\to 0^+}\frac{1/x}{-\left(x^x-1\right)^{-2}x^x(1+\ln x)}\\\\ &=-\lim_{x\to 0^+}\frac{\left(x^x-1\right)^2}{x(1+\ln x)}\\\\ &=-\lim_{x\to 0^+}\frac{2\left(x^x-1\right)(1+\ln x)}{2+\ln x}\\\\ &=-2\left(\lim_{x\to 0^+}(x^x-1)\right)\left(\lim_{x\to 0^+}\frac{1+\ln x}{2+\ln x}\right)\\\\ &=-2\cdot0\cdot1\\\\&=0\;, \end{align*}$$

and the desired limit is $1$.

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1  
Maybe the last equality should be $...=-2\cdot 0\cdot 1=0$ instead? –  Thomas E. Jun 5 '12 at 4:19
    
@Thomas: It should indeed; thanks. –  Brian M. Scott Jun 5 '12 at 5:10
    
Excellent! Alas, I went for the "$\frac{0}{0}$" version instead of the "$\frac{\infty}{\infty}$" version. –  Cameron Buie Jun 5 '12 at 16:15

All we need is a nice enough series expansion for $x^x$ about $0$, which can be obtained by rewriting $x^x$ as $\exp \left( x \log (x)\right)$ and looking at the Taylor series of $\exp \left( y\right)$. Now if we look at $$f(x) = \log \left( \dfrac{x^{x^x}}{x} \right) = \log \left( x^{x^x}\right) - \log x = \log (x) \left(x^x - 1 \right) \\= \log (x) \left(-1 + \left(1 + x \log (x) + \dfrac{x^2\log^2(x)}{2!} + \dfrac{x^3\log^3(x)}{3!} + \dfrac{x^4\log^4(x)}{4!} + \cdots\right) \right)\\=x \log^2(x) \left( 1 + \dfrac{x\log(x)}{2!} + \dfrac{x^2\log^2(x)}{3!} + \dfrac{x^3\log^3(x)}{4!} + \cdots\right)$$ Hence, the limit of $f(x)$ as $x \to 0$ is $0$. Hence, $$\lim_{x \rightarrow 0} \dfrac{x^{x^x}}{x} = 1$$

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It is a bit of a misnomer to call that a "Taylor" series of $x^x$. –  Antonio Vargas Jun 5 '12 at 3:54
    
@AntonioVargas Right :). I will change it. –  user17762 Jun 5 '12 at 3:54
    
My point was that $\sum x^k \log^k(x)/k!$ is not a Taylor series at all. Taylor series are power series, and that is not a power series. –  Antonio Vargas Jun 5 '12 at 3:57
    
@AntonioVargas I am writing out the Taylor series of $\exp(y)$ where $y$ is the underlying variable. I have now edited it accordingly. –  user17762 Jun 5 '12 at 3:58
    
A series approach, of course! I need to remember that I have that in my bag of tricks. –  Cameron Buie Jun 5 '12 at 16:17

This really isn't bad. We have

$\displaystyle \frac{x^{x^x}}{x} = \large e^{\log(x^{x^x}) - \log(x)} = e^{\log(x)(e^{x\log(x)}-1) } = e^{\log(x)^{e^{x\log(x)}-1}} = x^{e^{x\log(x) - 1}} $

and so the continuity of $\exp(x)$ ensures that

$\displaystyle \lim_{x \to 0^+} \frac{x^{x^x}}{x} = \lim_{x \to 0^+} \large x^{e^{x\log(x) - 1}} = x^{e^{\lim_{...} (x\log(x) -1)}} = x^0 $

and so the limit is 1, as promised.

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This doesn't seem right. What happened to the $-1$ in the exponent in the last step? Also, you can't let $x \to 0$ in some parts of the expression while leaving some $x$'s as they are. –  Hans Lundmark Jun 5 '12 at 19:53
    
@HansLundmark I have a replacement lined up here. Feel free to comment; if it's good, I'll edit it in. –  Chris Jun 6 '12 at 0:33

Using the standard limits $\lim_{x \to 0^+} x \, (\ln x)^a = 0$ (for $a>0$) and $\lim_{t \to 0} \frac{e^t-1}{t} = 1$ we find that $$ \frac{x^{x^x}}{x} = \frac{e^{x^x \ln x}}{e^{\ln x}} = e^{(x^x-1) \ln x} = \exp((e^{x \ln x}-1) \ln x) = \exp\left( \frac{e^{x \ln x} - 1}{x \ln x} \cdot x \,(\ln x)^2 \right) \to \exp(1 \cdot 0) = 1 $$ as $x \to 0^+$.

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Very slick, Hans! –  Cameron Buie Jun 5 '12 at 16:19

For a fast/simple solution i'm going to resort to 2 elementary limits, namely $\lim_{x\to0} \frac{\ln(1+x)}{x}=1$
and $\lim_{x\to0^+} x^x=1$. Let's proceed with the proof:

$$\lim_{x\to 0^+}\cfrac{x^{x^x}}{x}=\lim_{x\to 0^+}x^{x^x-1}=\lim_{x\to 0^+}x^{{\frac{x^x-1}{\ln x^x}}\cdot\ln x^x}=\lim_{x\to 0^+}x^{\ln x^x}=\lim_{x\to 0^+}e^{x {(\ln x)}^{2}}=1.$$

Q.E.D.

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