 Quantum reflection at a downward step

For $V_0<0$, $k=\sqrt{2mE}/\hslash$ tends to zero as $E\to 0^{+}$ while $q=\sqrt{2m(E+|V_0|)}/\hslash$ stays finite. Matching gives probability amplitudes $r=(k-q)/(k+q)$ and $t=2k/(k+q)$, so quantum reflection probability $R=r^2\to 1$ and transmission probability $T=(q/k)|t{|}^2\sim 4k/q\to 0$. The final region is classically allowed; reflection results from the abrupt spatial change in the wavefunction's wavenumber. These are limits of positive-energy incoming states, not a flux ratio evaluated at exactly zero incident current.