# Tetration convergence: prove $\lim_{x\rightarrow0} {}^{n}x = \begin{cases} 1, & n \text{ even} \\ 0, & n \text{ odd} \end{cases}$

I'm a computer student, learning math just for fun. Today I was graphing for fun that I found something strange! I noticed that that wired function $${x^{x^{\cdot^{\cdot^{x}}}}}$$ in zero, seems to converge to 1 when there are even powers and to 0 when there are odd powers! Then I attempt to prove it but I failed.

Then I did a little research and I found the Tetration article on Wikipedia. This article says that my guess was right but without any proof. So I'm here to ask you about it.

If we define $${x^{x^{\cdot^{\cdot^{x}}}}}$$ as $${^{n}x} :=\begin{cases} 1 &\text{if }n=0 \\ x^{\left(^{(n-1)}x\right)} &\text{if }n>0 \end{cases}$$

then prove:

$$\lim_{x\rightarrow0} {}^{n}x = \begin{cases} 1, & n \text{ even} \\ 0, & n \text{ odd} \end{cases}$$

• I believe you should be able to do this by induction straight from the definition. – Brevan Ellefsen Nov 9 '19 at 23:36
• @BrevanEllefsen I don't see a simple way to apply induction... the $0^0$ form is indeterminate. – Don Thousand Nov 9 '19 at 23:44
• @BrevanEllefsen I mean the induction step. The base cases are trivial. – Don Thousand Nov 9 '19 at 23:46
• @DonThousand use strong induction, consider whether your case is even or odd, propagate top to bottom until it is reduced to one of the two base cases. I hesitate to write more since I want the OP to have a chance to do this. – Brevan Ellefsen Nov 9 '19 at 23:47

As mentioned in the comments, you want to approach this by induction. What is missing in to show that as $$x\to 0^+, 1^-$$, then $$x^x$$ approaches $$1$$ from below.

Step 1: \begin{align}\lim_{x\to 0}x^x&=\lim_{x\to 0}e^{x\ln x}\\&=e^{\lim_{x\to 0}x\ln x}\\&=\exp\left({\lim_{x\to 0}\frac{\ln x}{1/x}}\right)\\&=\exp\left(\lim_{x\to 0}\frac{1/x}{-1/x^2}\right)\\&=e^0\\&=1 \end{align}

Step 2: \begin{align}\frac d{dx}x^x&=\frac d{dx}e^{x\ln x}\\&=e^{x\ln x}\frac d{dx}(x \ln x)\\&=x^x(\ln x+1) \end{align}

Step 3:

Using step 1 above, the limit of the derivative of $$x^x$$ when $$x\to 0^+$$ is $$-\infty$$. But the only thing we care is that it is negative

Step 4:

If $$1/e then the derivative of $$x^x$$ will be positive. At $$x=1$$, $$x^x=1$$, so for $$x$$ slightly less than $$1$$, $$x^x$$ is increasing towards $$1$$.

• Nice! Isn't step 3 superfluous, though? $x^x=e^{x\log(x)}$ is continuous at $x=1$, and $1^1=1$, so... – URL Nov 10 '19 at 6:46
• @URL It's important that $x^x$ is increasing towards $1$ when close to $1$, and decreasing when close to $0$. You need that for induction. If you increase $n$ by $2$ you get closer to $0$ or to $1$, depending on $n$. – Andrei Nov 10 '19 at 7:56

Note that:

$$x^x=\exp(x\ln(x))=1+\mathcal O(x\ln(x))\tag{x\to0}$$

and in general,

\begin{align}x^{1+\mathcal O(x\ln(x))}&=x\cdot x^{\mathcal O(x\ln(x))}\\&=x\exp(\mathcal O(x\ln^2(x)))\\&\sim x\tag{x\to0}\end{align}

and

\begin{align}x^{\mathcal O(x)}&=\exp(\mathcal O(x\ln(x)))\\&=1+\mathcal O(x\ln(x))\tag{x\to0}\end{align}

from which you can easily see that it will alternate between being approximately $$x$$ and $$1+\mathcal O(x\ln(x))$$, and hence the limit alternates between $$0$$ and $$1$$.

Sorry, I'm leaving an answer so I can show an image. Note the red colored large kidney in the center of the image. It is the location of period one convergence and is referred to as the Shell-Thron Region (STR). Immediately to left of the center of the STR is a small yellow disk of period two convergence. Note that $$^{\infty}1 = 1$$ is at the center of the STR where $$^{\infty}a = a$$, while center of the yellow disk is $$0$$.

A pragmatic answer is that $$1$$ drives the dynamics of the surrounding exponential Mandelbrot map and that $$0$$ does the same, therefore $$0^0=1$$. In combinatorics it is common to take $$0^0$$ as $$1$$.