Since we have made a transformation to spherical coordinates it means that $r\in [0,\infty)$. The quantum number $l$ is the orbital momentum of the electron. Then we substitute $R(r) = (1/r) u(r)$ and obtain $$-\frac{\hbar^2}{2 m} \frac{d^2}{dr^2} u(r) + \left ( V(r) + \frac{l (l + 1)}{r^2}\frac{\hbar^2}{2 m} \right ) u(r) = E u(r) .$$ The boundary conditions are $u(0)=0$ and $u(\infty)=0$.