Generalized recursive operators

[bo198214]

I think that we should have: [n -> + oo] lim (b[n](-k)) = -k + 1.

Ya thats guarantied already for the (unique) integer version. We have
b[n](-k)=-k+1 for 0\( \le \)k\( \le \)n-3. If n goes to \( \infty \) the k-range for b[n](-k)=-k+1 increases and in the limit reaches all positive integer k, i.e. \( \lim_{n\to\infty} \)b[n](-k) = -k + 1 for all k\( \ge \)0.

But my question was about whether the limits \( \lim_{x\rightarrow-\infty} \)b [2n+1] x depend on the concrete extension of the hyper operation to the reals (for example by Andrew's method, by regular iteration at some fixed point, by matrix operator method, by Jay's method, by linear/polynomial/exponential base function, etc.).

For example b[5](-k)=\( \text{slog}_b^{\circ k}(1) \) depends surely on the choice of \( \text{slog} \), which we per default regard to be Andrew's. However perhaps it can be that *in the limit* the specific choice of \( \text{slog} \) does not matter.