Confused regarding the special case "Wide deep well" of the Finite Square Well Potential in Griffiths book I am assuming that some of the equations I write are known to you people as they are all from Griffiths. Let me know if something seems vague.
So after solving the time-independent Schrödinger for the Finite square well potential - 
$$V(x)=\begin{cases}
-V_0  & \text{for}\,\, |x| \le a \\
\,\,\,\,\,0 & \text{for}\,\, |x|\gt a
\end{cases}$$
We end up with an equation whose LHS is dependent on $E$ and RHS dependent on $V_0$. I am of course talking about this equation: 
$$\tan z  = \sqrt{(z_0/z)^2 -1}\tag1$$
$$z=la$$
where $l = \frac{\sqrt{2m(E+V_0)}}{h}$ (Spring constant equivalent while solving Time independent Schrödinger for $|x| < a$)
$$z_0 = \frac{a \sqrt {2mV_0}}{h}$$
To realize the values of $E$, he draws the LHS and RHS of eq $(1)$ in z-space and he analyzes the meeting points. 
Special Case: Wide Deep well
In this case $z_0$ is very high, which corresponds to (according to the text), high value of $V_0$, the meeting points will be close to $z_n = n\pi/2$ and argues that 
$$ E_n + V_0 = \frac{n^2\pi^2h^2}{2m(2a)^2}$$   (Almost)
I didn't understand the following in that argument: 


*

*He says “But $E+V_0$ is the energy above the bottom of the well” — What is the purpose of this statement? 

*The finite square well goes on to infinite square well as $V_0 \, \rightarrow \infty$.  How?


First post here! Let me know if I can improve the question in some way. Thanks!
 A: The idea here is that if you take a finite square well and increase its depth, the situation should (in some sense) approach the infinite square well that was described earlier in that same chapter of Griffiths.  If this is the case, then the energy levels for the finite square well should approach the energy levels for the infinite square well as $V_0$ gets larger and larger.
What Griffiths argues is that when $V_0$ is large, the lowest energy levels satisfy (approximately)
$$
E_n \approx \frac{n^2 \pi^2 \hbar^2}{2m(2a)^2} - V_0.
$$
The energy levels for the infinite square well, meanwhile, were found to be
$$
E_n = \frac{n^2 \pi^2 \hbar^2}{2m(2a)^2}.
$$
Why are these expressions different?  Well, remember that the total energy of a particle is only defined in relative terms;  we can always add or subtract a constant to the energy.  Griffith's statement is basically that if we take "the bottom of the well" as our "zero point" for the energy in both cases, then in each case these energy levels are the same "energy distance" above the bottom of the well, which is all that matters.
So the energy levels are equivalent, at least;  this is a necessary, but not sufficient, condition that the finite square well approaches the infinite square well in the $V_0 \to \infty$ limit.  Griffiths doesn't actually show that everything is completely equivalent;  some of this is left to the exercises, and some of this he just skims over.  (For example, you might try thinking about what happens to the finite-well wavefunctions $\psi(x)$ in the limit $V_0 \to \infty$, and see if you can convince yourself that this approaches the wavefunctions for the infinite square well.)
A: First, note that equation (1) is (after substitutions)
$$\tan\left(\frac{a\sqrt{2m(E+V_0)}}{\hbar}\right) = \sqrt{-\frac{E}{E+V_0}}$$
where $E < 0$ (bound states). Setting $E=-V_0 + \epsilon$ (where $\epsilon$ is the energy above the bottom of the well), this equation becomes
$$\tan\left(\frac{a\sqrt{2m\epsilon}}{\hbar}\right) = \sqrt{\frac{V_0}{\epsilon}-1}$$
Now, in the limit as $V_0\rightarrow\infty$, it must be that
$$\frac{a\sqrt{2m\epsilon}}{\hbar} = (2n + 1)\frac{\pi}{2}$$
Solving the $\epsilon$ yields
$$\epsilon_m = \frac{m^2\pi^2\hbar^2}{2m(2a)^2},\quad m = 2n + 1$$
($m$ is odd since Griffiths is solving for the even wavefunctions and leaves the odd wavefunctions for problem 2.29)

He says "But $E+V_0$ is the energy above the bottom of the well" -
  What is the purpose of this statement?

In the above, I'm interested in solving for $\epsilon$, the energy above the bottom the well, since I'm aiming to find the solutions in the limit of an infinitely deep well.

The finite square well goes on to infinite square well as $V_0\rightarrow\infty$. How?

Yes, the infinite square well seems different since the bottom of the canonical infinite square well is at the zero potential level whereas this problem has the bottom of the well going to negative $\infty$ potential.  But, to emphasize the above, if we solve for the eigenvalues of the energy above the bottom of the well, the potential at the bottom of the well is not a factor.
