# what does $\lim _{x\to 0}$ $x^{x^{x^{x^{x^{x\cdots}}}}}$ evaluate to? [duplicate]

I was wondering if there is any possible way to define a limit for this form?

I tried using L’hopital rule and tried evaluating the limit:

$$\lim _{x\to 0} x^{x^{x^{x^{x^{x\cdots}}}}}$$

$$\lim _{x\to 0} x^{x^{x^{x^{x^{x\cdots}}}}} = \lim _{x\to 0} y$$

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

$$\lim _{x\to 0} \ln y = \lim _{x\to 0} \frac{\ln x}{\frac{1}{y}}$$

$$lim _{x\to 0}$$ $$\ln y$$ = $$lim _{x\to 0} \frac{\frac{1}{x}}{-\frac{1}{y^{2}}\frac{dy}{dx}}$$

$$lim _{x\to 0}$$ $$\ln y$$ = $$lim _{x\to 0}$$ $$-\frac{y^{2}}{x}\frac{dx}{dy}$$

$$lim _{x\to 0}$$ $$x^{x^{y}}$$ = $$lim _{x\to 0}$$ $$y$$

$$lim _{x\to 0}$$ $$\ln y$$ = $$lim _{x\to 0}$$ $$\ln\left(\frac{\ln y}{\ln x}\right)$$

following the same steps yield the result: $$lim _{x\to 0}$$ $$\ln\left(\frac{\ln y}{\ln x}\right)$$ = $$lim _{x\to 0}$$ $$-\frac{y^{2}}{x}\frac{dx}{dy}$$ substituting the previous result leads to the equation

$$lim _{x\to 0}$$ $$\ln y$$ = $$lim _{x\to 0}$$ $$\ln\left(\frac{\ln y}{\ln x}\right)$$

$$lim _{x\to 0}$$ $$x^{x^{x^{x^{x^{x...}}}}}$$ = $$lim _{x\to 0}$$ $$\log_{x}\left(x^{x^{x^{x^{x...}}}}\right)\$$

substituting $$x^{x^{x^{x^{x^{x...}}}}}$$ as 0 or 1 does not satisfy the equation (as manually checking the limit with a calculator yields a result of a number that approaches 0 or 1).

However if I let $$lim _{x\to 0}$$ $$x^{x^{x^{y}}}$$= $$lim _{x\to 0}$$ $$y$$ then I will get the result such that

$$lim _{x\to 0}$$ $$\ln\left(\frac{\ln\left(\frac{\ln y}{\ln x}\right)}{\ln x}\right)$$ = $$-\frac{y^{2}}{x}\frac{dx}{dy}$$ = $$lim _{x\to 0}$$ $$\ln y$$

therefore another equation is made such that $$lim _{x\to 0}$$ $$\log_{x}\left(\log_{x}\left(x^{x^{x^{x^{x...}}}}\right)\ \right)$$= $$lim _{x\to 0}$$ $$\ x^{x^{x^{x^{x...}}}}$$ However, substituting the value of $$\ x^{x^{x^{x^{x...}}}}$$ as 0 or 1 satisfies the above equation.

So how does this work?

• I have a guess that the limit would be equal to 1 – The 2nd Feb 25 at 4:01
• $x^{x^{x^\ldots}}$ is divergent by oscillation near $0$. We can see this because ${0}^{\varepsilon}=1$ but $0^1=0$. – Jam Feb 25 at 4:01
• It would be 1 ... – FishingCode Feb 25 at 4:02
• I cleaned up some of your MathJax code. That should suggest how to clean up the rest. – Michael Hardy Feb 25 at 4:04
• Please use \lim, when formatting, and not plain lim. – amWhy Feb 25 at 4:04

The infinite exponentiation converges for $$x\in [e^{-e},e^{1/e}]$$ which we can think of as the domain of the function we're taking limit of. Therefore, it's not clear how we can make sense of this limit as $$x\to 0$$. What you can do, instead, is compute $$\lim_{x\to 0^+}f_n(x)$$ where $$f_n(x)=x^{x^{x^{\cdot^{x}}}}$$ and we have $$x$$ appear $$n$$ times ($$n\in\mathbb{N}$$). It turns out the limit is $$1$$ when $$n$$ is even, and $$0$$ when $$n$$ is odd. See this answer for a proof and note the functions there are indexed slightly differently.
• @GottfriedHelms For $x\in(e^{-e},e^{1/e})$, if we define $f(t)=x^t$ and $a$ is a fixed point, i.e., $f(a)=a$, then we have $|f'(a)|<1$ and $f'$ is continuous. So those fixed points are attractive. When $x$ is equal to the boundary values of the interval, the situation is more complicated. – bjorn93 Feb 25 at 21:06
If you define $$f_1=x \qquad \text{and} \qquad f_n=x^{f_{n-1}}$$ and compose Taylor series around $$x=0$$, you should find that $$\forall n$$
$$f_{2n}=1+x \log(x)+\frac 12 x^2\log^2(x)(1+2\log(x))+O(x^3)$$ $$f_{2n+1}=x+x^2\log^2(x)+O(x^3)$$
$$\lim _{x\to 0} f_{2n}=1\qquad \text{and} \qquad \lim _{x\to 0} f_{2n+1}=0$$