 Solution

The delta-shell Born amplitude is first order in the potential strength:It retains finite-energy angular dependence and includes the leading weak-coupling contributions of all partial waves. By contrast, the supplied $-gR/(1+g)=-a_0$ is the zero-energy limit of the exact S-wave scattering amplitude. It is isotropic, depends nonlinearly on $g$, and captures the scattering-length pole missed by first-order perturbation theory.

At finite energy the exact S-wave amplitude is insteadwhich is generally complex and obeys elastic unitarity. Its zero-energy limit is $-a_0$ for finite scattering length. Thus the two expressions printed in the question represent different approximations; the energy-independent S-wave expression is not the full finite-$k$ result of part (c).

Their common regime requires both weak coupling and low energy:Then $qR\le 2kR\ll 1$, soConsequentlyin that joint regime. Low energy alone does not remove the nonperturbative denominator $1+g$, while weak coupling alone does not remove higher-partial-wave angular dependence.