Question: Define $f_0=x,f_{n+1}(x)=e^{1/f_n(x)}$ and $x\neq0$. What does this sequence of functions converge to as $n\to+\infty$? Is it true that $f_n(x)$ converges to the constant $\frac1{W(1)}$, where $W(x)$ is the Lambert W function?
Attempt: By plotting $f_n$ for high $n$, it seems to approach a constant function around $y\approx1.7$. By assuming that for sufficiently large $n$, $f_n=f_{n+1}$, and assuming that $f_n$ becomes constant, we can solve $f_n=e^{1/f_n}$ for $f_n=\frac{1}{W(1)}\approx1.763$.
Context: I was investigating the behaviour of $f_{n+1}(x)=g(f_n(x))$ for different $g(x)$ and found that $g(x)=e^{1/x}$ seemed to converge to a constant around $1.7$.
Notes: My attempt is by no means a rigorous proof. Is the limit of the sequence necessarily the solution of the functional equation $f(x)=e^{1/f(x)}$? And why should the limit be a constant? If anyone could direct me to a method for a proof or previous literature, I'd be appreciative. Thank you.
If the limit exists then it must be a fixed point of $f$ i.e. it must be some $x_0$ such that
$x_0 = e^{1 / x_0}$
Let $y_0=\frac{1}{x_0}$. Then
$\frac{1}{y_0} = e^{y_0}$
$\Rightarrow y_0 e^{y_0} = 1$
$\Rightarrow y_0=W(1)$
$\Rightarrow x_0 = \frac{1}{W(1)}$
So now you just need to prove that the limit exists. As a first step towards this, sketch the graph of $f(x)$ for $x \ne 0$.