Answer
See explanation.
Work Step by Step
It is true that as the depth of the finite well $U_o$ tends toward zero, the bound-state energy E is about $0.68U_o$ so it also tends to zero. This is discussed on page 1339. In physical terms, a very shallow energy well cannot confine particles that have too much energy. It makes sense that as the depth of the well approaches zero, the number of the particle’s trapped states decreases to just one, and its energy decreases.
This does not violate the Heisenberg uncertainty principle, because there is no state where the energy is exactly zero.
In mathematical terms, the wavefunction does not consist of a pure sine wave – it has exponentially decreasing tails outside – so it cannot be associated with a single value of k, nor a single value of the momentum p. In other words, there is a spread in its momentum $\Delta p$ even as the energy $\Delta E$ decreases.
