Question: I stumbled across an exercise in quantum mechanics concerning the self-adjointess of the radial momentum operator. In the exercise, the statement was made that this operator is self-adjoint for wave functions that satisfy

$$\lim\_{r\to0}r\psi\left(r\right)=\lim\_{r\to \infty}r\psi\left(r\right)=0$$

I was wondering, from a pure mathematical point of view, what an example of such a function $\psi\left(r\right)$ would be.

Answer: 
How about $\psi(r) = \frac{1}{\sqrt{2\pi}\; r^2} \exp(-1/r)$?

It satisfies both of your limits, and also is normalized in the sense that
$$
\int\_0^\infty dr\, 4\pi r\; {\left|\psi(r)\right|}^2 = 1\, .
$$

Something like $\frac{8^{1/4}}{{\pi}^{3/4}} \exp(-r^2)$ also works in this regard. Come to think of it, any function that is a polynomial in $r$ times $\exp(-a r^2)$ will work, where $a>0$. I believe this class would include the radial part of the eigenstates of the hydrogen atom.
$1/(a^2+r^2)$ will work for $a>0$; the normalisation constant is an exercise.