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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?

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

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  • $\begingroup$ What about the question whether the fixpoint is of repelling or of attracting type? (Likely it is easy but I've not been in such questions lastly) $\endgroup$ – Gottfried Helms Feb 25 at 10:41
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    $\begingroup$ @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. $\endgroup$ – bjorn93 Feb 25 at 21:06
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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 $\forall n$

$$f_{2n}=1+x \log(x)+\frac 12 x^2\log^2(x)(1+2\log(x))+O(x^3)$$ $$f_{2n+1}=x+x^2\log^2(x)+O(x^3)$$

$$\lim _{x\to 0} f_{2n}=1\qquad \text{and} \qquad \lim _{x\to 0} f_{2n+1}=0$$

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The Mandelbrot of the exponential map shows that zero is at the center of a period two region (the circular orange region). So all of the points near zero are also period two. So 0 and 1 make sense as solutions.

Mandelbrot of the exponential map

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