I want to rewrite $x^{x^{x^{x^{x...}}}}$ as a limit in the form of $$\lim_{n\to \infty}f(n)$$ I have defined the recurrence relation $$a_x(n) = x^{a_x(n-1)}$$ and $$a_x(1) = x$$ So, I can write the limit as: $$\lim_{n \to \infty}a_x(n)$$ but since my main goal is to evaluate the limit for $x = \sqrt2$, i.e $\lim_{n \to \infty}a_{\sqrt2}(n)$, this notation won't help. Please add some hints as to how should I proceed with solving the limit too. The page where I found this says that one may set $A = x^{x^{x^{x^{x...}}}}$, then $x^A = A$, so $x = A^{1\over A}$, which implies that for $A = 2,4$, $x = \sqrt2$. I have trouble finding which one is correct (using a calculator, I found it is very close to 2 for $n = 10$), but the page says something about finding the limit and leaves it there for the reader. Any help would be appreciated.

  • $\begingroup$ If you just write $x^{x^x}$ usually $x^{(x^x)}$ is meant, and with this interpretation $x=\sqrt{2}$ shoudl quickly diverge to infinity. Do you mean $(x^x)^x$? $\endgroup$
    – quarague
    Dec 11, 2022 at 18:02
  • $\begingroup$ If there is a limit $L$ of $a_x(n),$ then you will have $L=x^L.$ But you have to show the limit exists. Aside: I recommend using $b_n(x)$ rather than $a_x(n).$ We more often use indexes as integers and function arguments as reals. Your choice isn't wrong, it just reads ... odd. $\endgroup$ Dec 11, 2022 at 18:04
  • $\begingroup$ @quarague Actually, it doesn't diverge. I understand why you'd guess it does, but it doesn't. $\endgroup$ Dec 11, 2022 at 18:05
  • $\begingroup$ There are other related, but different, questions on the site. For examples, math.stackexchange.com/q/1948778 , math.stackexchange.com/q/4045353 , and math.stackexchange.com/q/1566845 . $\endgroup$
    – Xander Henderson
    Dec 11, 2022 at 18:05
  • 3
    $\begingroup$ @quarague For a quick argument why that isn't the case, note $(\sqrt{2})^x$ is an increasing function of $x$, so whenever $x < 2$ you have $(\sqrt{2})^x < (\sqrt{2})^2 = 2$. Now induct on the height of any finite tower: $\sqrt{2} < 2$, so $\sqrt{2}^\sqrt{2} < 2$, so $\sqrt{2}^{\sqrt{2}^\sqrt{2}} < 2$, and so on, showing every such tower of finite height is less than $2$ (definitely not diverging) $\endgroup$ Dec 11, 2022 at 18:58