Let $a$ be a real number with $a > e^{1/e}$ and $a <> e$.
$slog$ means superlog base $e$ and $sexp$ means superexp base $e$.
$sloga$ means superlog with base $a$ and $sexpa$ means superexp base $a$.
$k$ is a real number with $k>0$ and $x$ is a real number with $x>0$.
Conjecture
There exists a real constant $Q > 0$ for each k (as a function of k) such that as $x$ goes to +infinity :
$Q = \dfrac{sexp( slog(sexp(slog(x)+k)/sexpa(sloga(x)+k)) -k)}{x}$
or equivalently
$sexp(slog(Qx) + k) = sexp(slog(x)+k) / sexpa(sloga(x)+k)$
Now this is a conjecture about when $x$ goes to $oo$. If $k$ goes to OO we can show there is a $Q$ for that $k$. To see that we use the famous change of base :
lim $k$ (change of base)
$Ax = sexp( slog( sexpa(sloga(x)+k) ) -k)$
Where $A$ is a nonzero real number.
or equivalently
$sexp(slog(Ax) + k) = sexpa(sloga(x)+k)$
Now if we plug this in the previous equations we get
$=> sexp(slog(Qx) + k) = sexp(slog(x)+k) / sexp(slog(Ax)+k)$
And thus $Q$ and $A$ are trivially linked.
If $A > 1 \implies Q < 1$. If $A < 1 \implies Q > 1$ And $0<QA<oo$
Thus if $k$ goes to oo we have solved the Conjecture , but if $k$ is finite the use of the base change is more dubious.
How to prove the Conjecture ?
\infty
for $\infty$\dfrac{A}{B}
for $\frac {A}{B}$\text{slog}
for $\text{slog}$\text{slog}_a
for $\text{slog}_a$\text{sexp}
for $\text{sexp}$\text{sexp}_a
for $\text{sexp}_a$\geq
for $\geq$\implies
for $\implies$\left( {\large{\text{Parethesis that fit stuff}}}\right)
for $\left( {\large{\text{Parethesis that fit stuff}}}\right)$ &c $\endgroup$\operatorname
over\text
since it naturally handles spacing issues. $\endgroup$