Put c=mκ/ℏ2, kn=(n+1/2)π/a, and δn=a(λn−kn). The quantization equation gives δn=arctan(c/λn), so λnδn→c and δn→0. Therefore λn2−kn2→2c/a, proving the displayed absolute energy shift. Its ratio to the unperturbed energy tends to zero. The normalized unperturbed even state has probability density 1/a at the origin, so its first-order delta-barrier expectation is also κ/a.