= Solution
Define $C=2\pi G\rho_c$ and $\ell_f=[D/(g\Delta\rho)]^{1/4}$. Then
$$
\frac{Z}{C}=1-\frac{e^{-kt}}{1+(k\ell_f)^4}.
$$
At long <wavelength>, the mass deficit at the lower interface nearly cancels the surface excess, and the gravity field does not resolve their vertical separation. For $t>0$,
$$
\boxed{Z\sim Ckt\longrightarrow0\quad(k\to0).}
$$
The first mechanical contribution is fourth order in $k$, although the exponential also contributes terms at second and third order. If the ideal interface depth were zero, the leading term would instead be $C(k\ell_f)^4$.
At short <wavelength>, bending resists the interface's deflection, and any remaining buried signal is attenuated. Consequently
$$
\boxed{Z\longrightarrow C=2\pi G\rho_c\quad(k\to\infty).}
$$
This is the uncompensated surface-sheet value. For nonnegative $D,t$ the approach is from below. With $\beta=D/(g\Delta\rho)$,
$$
\frac1C\frac{dZ}{dk}
=e^{-kt}\left[\frac{t}{1+\beta k^4}
+\frac{4\beta k^3}{(1+\beta k^4)^2}\right]\ge0.
$$
The scales $kt\sim1$ and $k\ell_f\sim1$ respectively mark gravitational depth filtering and flexural resistance; which appears first depends on $t/\ell_f$. Admittance has dimensions of inverse time squared. The formal $k\to\infty$ limit describes this model, not an assertion that thin-plate theory remains accurate at arbitrarily short <wavelength>.
\Image[/past-exam-of-the-physics-of-the-earth-as-a-planet-course-of-the-university-of-cambridge/2015/paper-1-flexural-admittance.png]
{title=Compensated long wavelengths, flexural transition and the uncompensated gravity limit}
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