Limit involving tetration Let the notation be $a^{\wedge\wedge}b 
= \underbrace{a^{a^{\cdot^{\cdot^{a}}}}}_{b\,times}$ for tetration.
My mentor conjectured the following:

Let $n$ be a positive integer, then let $A(n)$ be any function that satisfies
$$\lim\limits_{n\to\infty} \left(e^{\frac1e} +
\frac1n\right)^{\wedge\wedge}\left[(10 n)^{1/2} + n^{A(n)} + C+o(1)\right] - n = 0$$
where $C $ is a constant. Then $\lim\limits_{n\to\infty} A(n) = \frac1e $

Conjectured by
tommy1729
Here : 
http://math.eretrandre.org/tetrationforum/showthread.php?tid=262&page=4
So could this be true ?
I have no idea how to do limits like this. I assume these type of limits are not in the books.
 A: Here is the correct formula, where $\eta=\exp(1/e)$ and $\alpha(x)$ is the upper repelling fixed point Abel function for iterating $x \mapsto \exp(x)-1$, which is generated using Ecalle's fps solution.  For details, see posts by Will Jagy on $\alpha(x)$: How to obtain $f(x)$, if it is known that $f(f(x))=x^2+x$? 
$$\lim\limits_{n\to\infty} \text{sexp}_{(\eta+1/n)}\left[\pi\sqrt{\frac{2\eta\cdot n}{e}} -2\right] \approx 388.7874$$
I obtained the above limit value numerically, and then used that result to generate a corrected equation for the Op's question.
$$\lim\limits_{n\to\infty} \text{sexp}_{(\eta+1/n)}\left[\pi\sqrt{\frac{2\eta\cdot n}{e}} + \alpha(\frac{n}{e}-1) + C\right] -n = 0$$
$$C \approx -2 - \alpha(\frac{388.7874}{e}-1)$$
I was about to post a closely related question about Pi in the Mandelbrot set; it takes about $\pi \sqrt{n}$ iterations to escape near the parabolic cusp at c=0.25+1/n.  Then I found this paper about the occurrence of Pi in the Mandelbrot set; although I haven't finished reading their paper, but presumably the same linear differential equation mechanisms can be used to justify the result, that it takes $\pi \sqrt{2n}$ iterations to "escape" for iterating $x \mapsto \exp(x)-1+\frac1n$, where we start at $x=-1+\frac1n$.  Both iterations involve perturbations of $\frac1n$ near a parabolic fixed point.  
http://www.doc.ic.ac.uk/~jb/teaching/jmc/pi-in-mandelbrot.pdf
https://people.math.osu.edu/edgar.2/piand.html
It is simpler and mathematically equivalent to work with iterating $f(x)=\exp(x)-1+\frac1n$.  Using the paper's methods, then one would want to prove it takes $\pi\sqrt{2n}$ iterations, for the function to begin growing, where growth would be defined as $f^{\circ \pi\sqrt{2n}}>2$; after that growth is superexponential.  
n for iterating $f(x)=\exp(x)-1+\frac1n$ is equivalent to $n=\ln(\ln(\eta+1/m))+1\approx \frac{e}{\eta\cdot m} + \frac{\mathcal{O}}{m^2}$ for iterating $g(x) =(\eta+1/m)^x$  Then there is a simple linear conversion $f^{\circ k} = \frac{g^{\circ k}}{e}-1$  That is why I used $\alpha(\frac{n}{e}-1)$  in my corrected solution equation for the Op's question.
