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What is: $$\lim_{x→0} x^{x^x}$$

I'm getting 0 as an answer, but I also got infinity as an answer. How would one solve this?

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    $\begingroup$ That won't be defined for $x<0,$ so best to write $\lim_{x\to 0^+}.$ $\endgroup$
    – zhw.
    Apr 19, 2015 at 3:35
  • $\begingroup$ You can only consider $x\to 0^+$. In that case, the limit is indeed $0$: $x^x$ is quite close to $1$ for small positive $x$, so your expression behaves approximately like $x^1$ for small positive $x$ $\endgroup$
    – MPW
    Apr 19, 2015 at 3:35
  • $\begingroup$ I do not think that incorrect answers should be regarded as correct, because it is rare on this site. The limit does not exist because the convergence is not defined before $e^{-e}$ as you can see on many Web sites including mathworld.wolfram.com/PowerTower.html $\endgroup$
    – user195934
    Feb 10, 2016 at 13:30

5 Answers 5

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If both limits exist, and the result isn't an indeterminate form, $\lim_{n\to\infty} x_n^{y_n} = \left( \lim_{n\to\infty} x_n \right)^\left(\lim_{n\to\infty} y_n\right)$.

In particular, $\lim_{x\to0^+} x^{x^x} = \left( \lim_{x\to0^+} x \right)^\left(\lim_{x\to0^+} x^x\right) = 0^1 = 0$.

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  • $\begingroup$ Why are you allowed to split one limit into two seperate limits? Usually this needs some sort of rigor justification. $\endgroup$
    – Imago
    Feb 10, 2016 at 11:54
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    $\begingroup$ @Imago $x^y$ is a continuous function of two variables for $x\geq 0,y>0$, so if $\lim x=x_0,\lim y=y_0$, then $\lim x^y=x_0^{y_0}$. $\endgroup$
    – Wojowu
    Feb 10, 2016 at 11:55
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Let $y = x^{x^x}$ then $\ln y = x^{x}\ln x $.
Since $ \lim_{x\to0^+} x^x = 1 $
We have $$ \lim_{x\to0^+} \ln y = \lim_{x\to0^+} x^{x}\ln x = -\infty $$ Thus $$ \lim_{x\to0^+} y = \lim_{x\to0^+} e^{\ln y} = 0 $$

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Using Taylor series around $x=0^+$ $$x^{x^x}=x+x^2 \log ^2(x)+\frac{1}{2} x^3 \left(\log ^4(x)+\log ^3(x)\right)+O\left(x^4\right)$$ then the limit is effectively $0$.

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I won't be giving a rigorous answer here, I'd just be giving a rough solution:

$$\lim_{x \to 0} x^{x^x} = \, ?$$

What if we have $\lim_{x \to 0} x^x$ what do we get?

$$\lim_{x \to 0} x^x = 0^0 = 1$$

So then we'd have $\lim_{x \to 0} x^1$:

$$\lim_{x \to 0} x^1 = 0^1 = 0$$

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  • $\begingroup$ Look what Eugene Shvarts has written above. $\endgroup$
    – Leon
    Feb 10, 2016 at 12:19
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Definition of a Tetration

For any positive real $x\gt 0$ and non-negative integer $n$, we have $$\underbrace{x^{x^{\cdot^{\cdot^x}}}}_{n}=\ ^n x=\begin{cases} 1, & n=0\\ x^{\ ^{(n-1)} x}, & n\gt 0\\ \end{cases}$$

The $n^{\rm{th}}$ tetration of $0$ is not consistently defined. However, the limit of the $n^{\rm{th}}$ tetration of $x$ as $x$ approaches zero from the right is well defined. In general we have

$$\lim\limits_{x\to 0^+} \underbrace{x^{x^{\cdot^{\cdot^x}}}}_{n}=\lim\limits_{x\to 0^+} \ ^n x= \begin{cases} 1, & n\ \mbox{is even}\\ 0, & n\ \mbox{is odd}\\ \end{cases}$$

Therefore $$\lim\limits_{x\to 0^+} x^{x^x}=\lim\limits_{x\to 0^+} \ ^3 x=0$$

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