It is well-known that the finite square well always has a bound state, no matter how shallow it is. Suppose the finite square well is given by $V(x) = -V_0$ for $-a\le x \le a$. The standard result is that the ground state can be obtained by solving the following transcendental equation in $z$:
$$ z \tan z = \sqrt{z_0^2 - z^2}$$
where $z_0 := \frac{a}{\hbar}\sqrt{2mV_0}$ and the ground state energy can be calculated as
$$E = -V_0 \left(1-\frac{z^2}{z_0^2}\right)$$
My problem concerns shallow wells, i.e. the limit where $z_0\to 0$. I read a claim that the ground state energy $E$ can be estimated by $-z_0^2 V_0$ but I am having trouble proving this. I know some form of series expansion and truncating the higher order terms should take place, but I don't see how to approach the transcendental equation with this idea (since both $z$ and $z_0$ will be small).
Is this estimate correct? If not, what is the estimated energy in terms of $V_0$ and $z_0$ in this limit?
