Q:

Is $e^{-r/2}$ equivalent to $r^{-(l+1)}$ in the radial solution of Laplace equation? 

When we solve the Laplace equation for Hydrogen Wave Equation at large r, we obtain the expression below to account for the behavior of the wave at very very large $r$
$$R=e^{-(r/2)}$$
At very small $r$ we will have a general solution for the radial equation to be
$$R=Ar^l+Br^{-(l+1)}$$
Well interestingly, when I am learning DE solution for Laplace, the second part in the equation above usually takes charge of very large $r$. If that is the case, could $Br^{-(l+1)}$ be equivalent to $e^{-(r/2)}$?  If not, when would be the best case to use one from the other? 

A:

The solution for the radial part of the Schroedinger equation is:$$
R_{n,l}(r)=\rho^lL_{n,l}(\rho)e^{-\rho/2}
$$
where $\rho \propto r$ and $L_{n,l}$ is a Laguerre polynomial. 
This equation is dominated by the exponential term for large $r$ but as $r$ aproaches $0$ it is the polynomial term that dominates the expression. 
That is why you have different approximations for the wave equation. 
Now, it is not easy to tell when you have to use one or another of the two expressions.  What you can do is to verify that the exponential factor is little enough so that you can assume it is almost $1$: $$
e^{-\rho/2} \gt(1-\epsilon) \iff \rho/2\lt\epsilon \iff r \lt \frac{2\epsilon}{\sqrt{-\frac{2m_e}{\hbar^2}E}} 
$$
Take $\epsilon=.1$ or less and you will know if the polynomial approxiamtion is yet valid.