# Limit $\lim_{x\to 0} x^{x^x}$

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?

• That won't be defined for $x<0,$ so best to write $\lim_{x\to 0^+}.$
– zhw.
Apr 19, 2015 at 3:35
• 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 $x$
– MPW
Apr 19, 2015 at 3:35
• 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
– user195934
Feb 10, 2016 at 13:30

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

• Why are you allowed to split one limit into two seperate limits? Usually this needs some sort of rigor justification. Feb 10, 2016 at 11:54
• @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 $\lim x^y=x_0^{y_0}$. Feb 10, 2016 at 11:55

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 $$\lim_{x\to0^+} y = \lim_{x\to0^+} e^{\ln y} = 0$$

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 $0$.

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

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

• Look what Eugene Shvarts has written above.
– Leon
Feb 10, 2016 at 12:19

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