Evaluating negative infinite tetration: $\lim\limits_{n\to-\infty}\,^n x=\lim_{n\to\infty}\underbrace{\log_x\left(…\log_x(0)\right)}_n=\,^{-\infty} x$ One can learn that for a power tower of height $n$:
$$y=n\big\{x^{x^{x^…}}=\,^nx\implies \log_x(y)=\boxed{\log_x(\,^nx)=\,^{n-1}x}$$
giving a recursive relation. One might see that $n<-2$ cases are undefined, but using the extended real numbers or limits, one can actually use the principal branch for the logarithm function for $x\ne0$:
$$\log_x(\,^1x)=\frac{\ln(\,^1 x)}{\ln(x)}=\,^{1-1}x=\,^0x=1\implies  \frac{\ln(1)}{\ln(x)}=\,^{0-1}x=\,^{-1}x =0$$
Here is when problems occur. We can define $\ln(-\infty)=\infty$ or $\ln(-\infty)=\ln(-1)+\ln(\infty)=\pi i+\infty$, but to stick with real values, like in

Does this limit exist or is undefined $$\lim_{x\to -\infty}\ln\left(\frac{x^2+1}{x-3}\right)=\ln(-\infty)=\infty,$$

let us consider the simpler version of $\boxed{\ln(-\infty)=\infty}$. Note that I will not simplify the logarithms into the infinity to keep the domain as large as possible:
$$\frac{\ln(\,^{-1} x)}{\ln(x)}=\frac{\ln(0)}{\ln(x)}=\,^{{-1}-1}x=\,^{-2}x =-\frac{\infty}{\ln(x)}\\\implies \frac{\ln(\,^{-2} x)}{\ln(x)}= \frac{\ln\left(\frac{-\infty}{\ln(x)}\right)}{\ln(x)}=\,^{{-2}-1}x=\,^{-3}x = \frac{\ln\left(\frac{-\infty}{\ln(x)}\right)}{\ln(x)} = \frac{\infty-\ln(\ln(x))}{\ln(x)}\\\implies \frac{\ln(\,^{-3} x)}{\ln(x)}=\frac{\ln\left(\frac{\infty-\ln(\ln(x))}{\ln(x)}\right)}{\ln(x)}=\,^{{-3}-1}x=\,^{-4}x = \frac{\ln\left(\frac{\infty-\ln(\ln(x))}{\ln(x)}\right)}{\ln(x)}=\log_x\left(\infty-\ln(\ln(x))\right)-\log_x(\ln(x)) $$
and so on. This creates a couple questions using our aforementioned assumptions and these Wolfram Functions conventions. The question here is what would happen if we calculated:
$$\lim_{n\to-\infty}\,^n x=\lim_{n\to\infty}\underbrace{\log_x\left(…\log_x(0)\right)}_n=\,^{-\infty} x$$
which looks like a W-Lambert function:
$$-\frac{\text W_{-1}(-\ln(x))}{\ln(x)}= \lim_{n\to\infty}\underbrace{\log_x\left(…\log_x(e)\right)}_n $$
using the same process and the recursive definition of:
$$\log_x(\,^nx)=\frac{\ln(\,^nx)}{\ln(x)}=\,^{n-1}x $$ Also, what is the range of this new experimental $\ \,^{-\infty} x\ $ function? I have listed the conventions that will be used, so there should be no confusion. Please correct me and give me feedback!
 A: With some rough experimentation, one has the following result for the “experimental $-\infty$ tetration” using this computation and the sign function:
$$\log_{a+bi}\left(\log_{a+bi}\left(\log_{a+bi}\left(0\right)\right)\right)=\frac{\infty}{\text{sgn}(\ln(a+bi))}$$
It does not matter if you define $\ln(-\infty)=\infty+\pi i$ or $\ln(-\infty)=\infty$, for any amount $n>2$ logarithms nestings, you get the following:
$$\,^{-\infty}x\mathop=^\text{def}\lim_{n\to\infty}\underbrace{\log_x\left(…\log_x(0)\right)}_n= \frac{\infty}{\text{sgn}(\ln(x))}=\frac{\sqrt{\text{Re}^2(\ln(x))+\text{Im}^2(\ln(x))}}{\ln(x)}\infty$$
You may think that this is useless because it diverges, but the rate of divergence is proportional and is called the Directed Infinity function:
$$\text{DirectedInfinity}(x)\mathop=^\text{def} \infty x\implies \,^{-\infty}x = \text{DirectedInfinity}\left(\frac{1}{\text{sgn}(\ln(x))}\right) $$
which is a complex number times infinity.
One property of $\infty x$ is that:
$$\infty x=\infty \text{sgn}(x)\implies \infty\text{sgn}\left(\frac{1}{\text{sgn}(\ln(x))}\right)= \frac{\infty}{\text{sgn}(\ln(x))} $$
Here is a complex plot of $$\frac{1}{\text{sgn}(\ln(x+yi))} :$$

Here are some values of this function:
$$\frac{\infty}{\text{sgn}(\ln(2))}=\infty$$
$$\frac{\infty}{\text{sgn}(\ln(-2))}=\frac{\sqrt{\pi^2+\ln^2(2)}}{\ln(2)+\pi i}\infty$$
$$\frac{\infty}{\text{sgn}(\ln(i))}=-\infty i$$
$$\frac{\infty}{\text{sgn}(\ln(2))}=\infty$$
$$\frac{\infty}{\text{sgn}(\ln(2-3i))}=\frac{i\sqrt{\ln^2(13)+4\tan^{-1}\left(\frac32\right)^2}}{2\tan^{-1}\left(\frac32\right)+i\ln(13)}\infty$$
Please correct me and give me feedback!
