Let $x \uparrow \uparrow n = x^{x^{x^{x}}}$ ex: $x \uparrow \uparrow 3 = x^{x^x}$
I'm trying to evaluate the limit
$$ \lim_{n \rightarrow \infty} \frac{3 \uparrow \uparrow \frac{1}{n} - 1 }{2 \uparrow \uparrow \frac{1}{n} - 1 }$$
Obviously L'hopitals rule has some value here, but once applied i am stuck attempting to differentiate the tetration function and i'm not sure how to go about doing this.
This problem originates from attempting to find a homomorphism for exponentation. That is a complex function $g(x)$ such that $g(x^y) = g(x)^{g(y)}$
If $a\uparrow\uparrow(1/n)$ means the $n$th superroot of $a$, usually rendered $\sqrt[n]a_s$, then for $a\ge 1$ there are two possible behaviors as $n\to\infty$:
If $1\le a\le e$ there is a unique base $s$, with $1\le s\le\exp(1/e)$, for which $s\uparrow\uparrow n$ converges to the selected value of $a$, namely $s=a^{1/a}$.
If $a>e$, then the superroot is always greater than $e^{1/e}$ but the tetrations of any fixed number greater than $e^{1/e}$ will blow up past the target value of $a$. Therefore the $n$th superroot of $a$ is forced to converge to $e^{1/e}$ as $n\to\infty$.
So in the above example with $3>e$ but $2<e$, the limit is simply
$\dfrac{e^{1/e}-1}{\sqrt2-1}=(e^{1/e}-1)(\sqrt2+1)$.
Note that neither term of the fraction goes to zero because superroots do not in general converge to $1$. Therefore the form is not indeterminate.