It is shown that for singular potentials of the form lambda/r(alpha),t
he asymptotic form of the wave function both at r --> infinity and r -
-> 0 plays an important role. Using a wave function having the correct
asymptotic behavior for the potential lambda/r(4), it is, shown that
it gives the exact ground-state energy for this potential when lambda
--> 0, as given earlier by Harrell [Ann. Phys. (NY) 105, 379 (1977)].
For other values of the coupling parameter X, a trial basis;set of wav
e functions which also satisfy the correct boundary conditions at r --
> infinity and r --> 0 are used to find the ground-state energy of the
singular potential lambda/r(4) It is shown that the obtained eigenval
ues are in excellent agreement with their exact ones for a very large
range of lambda values.