Does $y=\lim_{n\to\infty}x\uparrow\uparrow n-f(x,n)$ have a root at $2$?

While messing around on Desmos calculator, I found an interesting factorial/tetration function, where we are using Knuth arrow notation.

$$y=\lim_{n\to\infty}x\uparrow\uparrow n-f(x,n)\\f(x,n)=x\underbrace{!\dots!}_n$$

Where there are $n$ factorials following $x$ in $f(x,n)$.

Looking at it graphically, you may find the link above, it appears there are two roots, $x=1,a$, where $a$ seems to get increasingly closer to $2$ as $n$ goes to infinity.

However, I can't figure out how to prove this because limits like this are not the sort I can deal with.

I do notice, however, that the following is the infinite power tower:

$$g(x)=x^{x^{x^{x^{\dots}}}}$$

And the inverse of the infinite power tower is:

$$g^{-1}(x)=\sqrt[x]x$$

So the real problem is probably in the my factorial function $f(x,n)$, which I find hard to deal with.

Any solutions or directions to point?

EDIT: It appears as though this root, $x=a$, will most likely occur higher than $2$, which is going to give us a problem. The infinite power tower doesn't hold for $x>\sqrt[e]e$, which means we might have to do this some harder way...

• It's well proven that the tetration of height $n$ is always greater than a string of factorials of length $n$, so that's a start.... at least you have to only approach the limit one way – Brevan Ellefsen Jan 8 '16 at 23:20
• @BrevanEllefsen Sadly, it's well proven for $x\ge3$, but that doesn't help. – Simply Beautiful Art Jan 8 '16 at 23:21
• That cannot be possible because the iterated factorial has a fixpoint at x=2 and from any point x=2+ eps going to eps=0 the value for y for already small n goes to (2+eps)^^n -(2+eps)!! and increases thus rapidly and there is no crossing of the real axis in the neighbourhood of x=2 (of course always the gamma function is used for factorials at fractional values) – Gottfried Helms Jan 14 '16 at 20:05
• (For small n) it seems the crossing of the real axis goes to something between 2.34 and 2.35 (I used x^^n/x(!^n) -1 and did n=5 using one step of logarithms) – Gottfried Helms Jan 14 '16 at 20:28