Is this a proper function and if so, has it a limit at infinity? I want to define a function implicitly by $$x^y = y^x$$ with $$x \ne y$$
such that $f: X \rightarrow Y$ is constrained by $X = \{ x \in \mathbb{R} | x > 1 \}$ and $Y = \{ y \in \mathbb{R} | y > 0 \}$. 
Is this a properly defined function? 
If so, has it a limit for $x \rightarrow \infty$?
I suspect it to approach $1$ and have tried to calculate some values numerically, but Wolframalpha calculates this only until around $x =40$.
 A: Taking logs on both sides yields
$$
\frac{x}{\log x} = \frac{y}{\log y}
$$
provided that neither of them is 1. In fact, the following function can be analyzed:
$$
f(t) = \frac{t}{\log t},\quad t \in (1,\infty)
$$
This function has a convex shape on the domain $(1,\infty)$, and its local minimum is at $t = e$. Now the domain and range of the function can be modified into $X = [e,\infty)$ and $Y = (1,e]$ so that the map is well-defined.
The limit is as suspected, 1.
A: Since $x,y\gt 0$ under consideration, we can use logarithms here.
$$x^y=y^x\implies y\ln x=x\ln y\implies \frac{\ln x}x=\frac{\ln y}y$$
Now, define the map $\varphi\colon t\mapsto\dfrac{\ln t}t$ for $t\gt 0$. Note that $\varphi'(t)=\dfrac{1-\ln t}{t^2}$, so $\varphi$ is strictly increasing on $[1,e)$ and strictly decreasing on $[e,\infty)$.
Our condition is $\varphi(x)=\varphi(y)$. Note that a solution $(x,y)$ to $x^y=y^x$, by symmetry, gives $(y,x)$ a solution as well.
Since the map $\varphi$ is strictly decreasing on $[e,\infty)$ while strictly increasing on $[1,e)$ and is continuous as well on $[1,\infty)$, the implicit equation $x^y=y^x$ does give an injective function $x\mapsto y=f(x)$ for $x\gt 1$ such that $f(x)\ne x$ except at $x=e$ and as $x\to\infty$, the function $f$ approaches the value attained by $f$ at $1$, ie, $1^y=y^1$, or $f(1)=1$
A: Yes, the function exists. A parametrization can be obtained by assuming that for some pair $(x,y)$ on the curve the relation $y = px$ holds. Substitution of this relation in the original expression yields:
$$x^{px} = (px)^{x}$$
Dividing both sides by $x^{x}$:
$$x^{(p-1)x} = p^{x}$$
Identifying that both sides are raised to the power of $x$ implies that:
$$x^{(p-1)}=p$$
And finally we get: $$x = p^{1/(p-1)}$$
Which implies that $y$ is given by: $$y = px = p^{p/(p-1)}$$
One can vary the parameter $p$ between zero and infinity. This yields a hyperbola shaped curve, symmetric around $y = x$. For $x$ to infinity indeed $y$ goes to $1$. The intersection of the curve with the line $y=x$ occurs in the point $(e,e)$. Given this result, you can choose or adjust the domain and range of the function.
