If $f(s) = (1+s)^{(1+s)^{(1+s)/s}/s}$, show that $\lim_{s \to \infty} f(s)/s = 1$. If $f(s) = (1+s)^{(1+s)^{(1+s)/s}/s}$, show that $\lim_{s \to \infty} f(s)/s = 1$.
This function comes up
in the parameterization
of the solutions to
$x^y = y^x$.
See for example, here:
Are there real solutions to $x^y = y^x = 3$ where $y \neq x$?
If we write
$y = rx$,
we get
$x = r^{1/(r-1)}$
and
$y=r^{r/(r-1)}$.
Then
$x^y
=(r^{1/(r-1)})^{r^{r/(r-1)}}
=r^{r^{r/(r-1)}/(r-1)}
$.
Finally,
if we write
$r = 1+s$,
this is
$f(s) = (1+s)^{(1+s)^{1+1/s}/s}
$.
Wolfy says that,
around $s=0$,
$$f(s)
=e^e +  \dfrac{e^{1 + e} s^2}{24} 
- \dfrac{e^{1 + e} s^3}{24} 
+ \dfrac{e^{1 + e} (219 + 5 e) s^4}{5760} + O(s^5)
$$
and,
around $s = \infty$,
$$f(s)
=s + (\log^2(1/s) - \log(1/s) + 1) 
+ \dfrac{\log^4(1/s) - 3 \log^3(1/s) + 5 \log^2(1/s) - 6 \log(1/s) + 2}{2 s} + O((1/s)^2),
$$
calling this a
generalized Puiseux series.
My question is:
How to show that that expansion
at $s = \infty$ is correct.
I would be satisfied with
a proof that,
as the title says,
$\lim_{s \to \infty} \dfrac{f(s)}{s}
= 1
$.
It might help
to note that
$\dfrac{y}{x}
= r$
and
$\dfrac{r}{s}
= \dfrac{1+s}{s}
= 1+\dfrac1{s}
$.
 A: The expansion at infinity is actually easy.  Following general methods to provide expansions at infinity:
$$f(1/s)=(1+1/s)^{(1+1/s)^{(1+1/s)/(1/s)}/(1/s)}=(1+1/s)^{(s+1)\cdot(1+1/s)^s}$$
As $s\to0^+$, we know that
$$(1+1/s)^s\to1$$
$$\implies(1+1/s)^{(s+1)\cdot(1+1/s)^s}\sim(1+1/s)^{(s+1)}$$
Which gives us
$$s\cdot f(1/s)\sim s\cdot(1+1/s)^{(s+1)}\to1$$
since
$$s\cdot(1+1/s)^{(s+1)}=\frac{s^{s+2}}{(s+1)^{s+1}}\to\frac11$$
coming from well known proofs of $\lim_{x\to0^+}x^x=1$.
A: $$\frac{f(s)}{s} = (1+s)^{(1+s)^{(1+s)/s}/s}/s$$
$$\log(\frac{f(s)}{s}) = \frac{(1+s)^{(1+s)/s}\log(1+s)}{s}-\log(s)$$
We are now claiming that
$$\lim_{s \to \infty} \left(\frac{(1+s)^{(1+s)/s}\log(1+s)}{s}-\log(s)\right)=0$$

We now see that the LHS can be changed into
$$ \lim_{s \to \infty}\left(\frac{(1+s)^{(1+s)/s}\log(1+s)-s\log(s)}{s}\right)$$
Now is as good a time as ever to let $r=s+1$
$$ =\lim_{r \to \infty}\left(\frac{r^{\frac{r}{r-1}}\log(r)-(r-1)\log(r-1)}{r-1}\right)$$
$$ =\lim_{r \to \infty}\left(\frac{r\log(r)-(r-1)\log(r-1)}{r-1}\right)$$
By L'Hopital
$$ =\lim_{r \to \infty}\left(\log(r)-\log(r-1)\right) = \color{red}{0}$$
