Classical limit of a quantum system If we have a one dimensional system where the potential 
$$V~=~\begin{cases}\infty & |x|\geq d, \\ a\delta(x) &|x|<d, \end{cases}$$ 
where $a,d >0$ are positive constants, what then is the corresponding classical case -- the approximate classical case when the quantum number is large/energy is high?
 A: Firstly, it's easy to start off with just the Dirac delta potential and see what that does. Wiki has a nice solution for the Delta fuction potential, and I am lifting off parts of it here.
Consider a potential $V(x) = a\delta (x)$ and consider a scattering like configuration, where a plane wave $e^{ikx}$ is incident from the left.
$$
\psi(x)=\begin{cases}e^{ikx}+re^{-ikx} & x<0 \\ te^{ikx} & x> 0\end{cases}
$$
By matching the boundary conditions, like on the wiki page, you get
$$
t = 1+r\\
(1-\alpha)t = 1-r
$$
where
$$
\alpha = \frac{ 2ma}{ik\hbar^2}
$$
characterizes the effect of the delta potential. Solving for $r$ and $t$, 
$$
t = \frac{1}{1-\alpha/2}\\
r=-\frac{\alpha/2}{1-\alpha/2}
$$
Now, it is easy to see that for high incident $k$, the only effect of the dirac delta potential is to write a phase discontinuity on the wavefuction. This is because, as $k$ increases, the transmission $|t|^2=1/(1+|\alpha|^2/4)$ approaches 1, but the transmitted wavefunction gets an extra phase given by 
$$
\text{Arg}(t) = -\tan^{-1}(|\alpha|/2)
$$
Getting back to the problem at hand, for a particle in a box (without the delta function), the allowed $k$ vectors are given by forcing the wavefunction to be zero at the walls at $x=-d$ and $x=d$, which gives us the condition
$$
k_n=\frac{\pi n}{2d}
$$
If now, we add a delta potential, then for high values of $n$ (or $k$), all the delta function will do is introduce a phase discontinuity at the origin, and consequently what you should expect is that the boundary condition is matched not for $k_n$, but something slightly off $k_n+\delta k_n$, where $\delta k_n$ is a small correction due to the delta function potential. For high values of $n$, this correction would drop, as the phase discontinuity decreases, and for classical like states (very large $n$) you expect to recover 1D box states, as mentioned by John Rennie.
A: Here we derive the bound state spectrum from scratch. Not surprisingly, the conclusion is that the Dirac delta potential doesn't matter in the semi-classical continuum limit, in accordance with Spot's answer.
The time-independent Schrödinger equation reads for positive $E>0$,
$$ -\frac{\hbar^2}{2m}\psi^{\prime\prime}(x) ~=~ (E-V(x))\psi(x), \qquad V(x)~:=~V_0\delta(x)+\infty \theta(|x|-d), \qquad V_0~>~0, $$
with the convention that $0\cdot \infty=0$.
Define
$$v(x) ~:=~ \frac{2mV(x)}{\hbar^2}, \qquad e~:=~\frac{2mE}{\hbar^2}~>~0 \qquad k~:=~\sqrt{e}~>~0\qquad 
v_0 ~:=~ \frac{2mV_0}{\hbar^2}. $$
Then 
$$ \psi^{\prime\prime}(x) ~=~ (v(x)-e)\psi(x). $$
We know that the wave function $\psi$ is continuous with boundary conditions 
$$\psi(x)~=0 \qquad  {\rm for}\qquad  |x|\geq d.$$ 
Also the derivative $\psi^{\prime}$ is continuous for $0<|x|<d$, and possibly has a kink at $x=0$,
$${\lim}_{\epsilon\to 0^+}[\psi^{\prime}(x)]^{x=\epsilon}_{x=-\epsilon} ~=~v_0\psi(x=0). $$
We get $$\psi_{\pm}(x)~=~A_{\pm}\sin(k(x\mp d))\qquad {\rm for } \qquad 
0 \leq \pm x \leq d.$$


*

*$\underline{\text{Case} ~\psi(x=0)=0}$. Then 
$$n~:=~\frac{kd}{\pi}~\in~ \mathbb{N}.$$ 
We get an odd wave function 
$$\psi_n(x)~\propto~\sin(kx).$$
In particularly, the odd wave functions do not feel the presence of the Dirac delta potential.

*$\underline{\text{Case} ~\psi(x=0)\neq 0}$. Then continuity at $x=0$ implies that the wave function is even $A_{+}+A_{-}=0$. Phrased equivalently,
$$\psi(x)~=~A\sin(k(|x|-d)).$$
The kink condition at $x=0$ becomes
$$ v_0A\sin(-kd)~=~2kA \cos(kd), $$
or equivalently,
$$ v_0\tan(kd)~=~-2k.$$
In the semiclassical continuum limit 
$$k \gg \frac{1}{d}, \qquad k \gg v_0,$$ 
this becomes 
$$\frac{kd}{\pi}+\frac{1}{2}~\in ~\mathbb{Z}, $$
i.e., in the semiclassical continuum limit the even wave functions do not feel the presence of the Dirac delta potential as well.
