tetration base conversion, and sexp/slog limit equations

[sheldonison]

I was thinking over the topic and try to put it on its feet.
The paradigm is that \( \lim_{x\to\infty}\text{slog}_a(\text{sexp_b}(x)) - x \) shall exist for each \( a,b>\eta \).
....
\( \lim_{n\to\infty}\log_b^{\circ n}({\exp_a}^{\circ n}(y)) =\text{sexp}_b(\text{slog}(y)-c_b) \)

I was able to follow up to this point, but shouldn't it be this, where slog(y) substitutes for x?
\( {\exp_a}^{\circ n}(\text{slog}y)))= \)

\( \kappa_{b,a}(y) = \lim_{n\to\infty} {\log_b}^{\circ n}({\exp_a}^{\circ n}(y)) \), \( \kappa_{a,b}={\kappa_{b,a}}^{-1} \)

sadly, I didn't understand this step.

.....
So we need to show in your case that the limit:
\( \lim_{b\to\eta^+}\text{slog}_b(\kappa_{b,a}(x)) \) exists for \( x \) in some initial range, where \( \text{slog}_b \) is your linear approximation.

To show this we could use the Cauchy criterion....

How is this different from:
\( \lim_{b\to\eta^+}\text{slog}_b(\text{sexp}_a(x)) \) exists ...
Is it because we don't know if sexp is an analytic function, but there are better reasons to believe that \( \kappa_{b,a} \) is an analytic function?

I have a plot of the wobble for sexp_version_a\( _e \)(x)- sexp_version_b\( _e \)(x) in the critical section, (-1 to 0), where sexp_version_a is derived from sexp\( _{1.45} \)
The difference is in the "odd" portion of the function; it is larger in the sexp derived from a base conversion from base 1.45