Connection between Schrödinger equations for a finite triangular well and a finite square well Suppose we solve the Schrödinger equation,
\begin{equation}
-\psi''(x) + V(x) \psi(x) = -|E| \psi(x),
\end{equation}
where two cases are considered, $V(x) = -V_0$ (square well) and $V(x) = -V_0 (1 - |x|)$ (triangular well) in the interval $[-1,1]$, and $V(x) = 0$ outside this interval. The solution is supposed to be zero in zero, $\psi(0) = 0$. In the first case, we have the trigonometric functions of the form $\sin(\sqrt{|E|}x)$, the second one gives the Airy functions of the form $\operatorname{Ai}(V_0^{-2/3}(V_0|x|-(V_0-|E|)))$.
My question is the following, is it possible to consider the square well as a limit of the triangular well by introducing a parameter $\varepsilon$ for the equation inside the interval $x \in [-1,1]$
\begin{equation}
-\psi''(x) - V_0 (1 - \varepsilon |x|)\psi(x) = -|E| \psi(x),
\end{equation}
taking the limit $\varepsilon \rightarrow 0$ in the solution and finishing with the trigonometric functions? The problem is such that the parametrized equation has the solutions of the form $\operatorname{Ai}(\varepsilon^{-2/3}(\varepsilon|x|-(1-|E|)))$, which do not converge to the usual trigonometric function but to the "decaying" one since the Airy functions converge to zero for big negative arguments. Where is the error?
 A: In fact, if we calculate the first eigenvalues and eigenfunctions, we will not find a big difference between the eigenfunctions, and the eigenvalues for large $n$ almost coincide. Put $V_0=1$, then the first 4 eigenfunctions for two types of potentials shown in Figure 1

The first 7 eigenvalues for two types of potentials:
1.4674, 8.86974, 21.2081, 38.4868, 60.7166, 87.9195, 120.134
2.16793, 9.37633, 21.7317, 38.9876, 61.2251, 88.4198, 120.638
The difference in eigenfunctions for the two types of potentials is shown in Fig. 2. It can be seen that at large values of $n$ the difference tends to zero, which is already obvious. 

A: To see what you've actually done here :
$$-\frac {d^2\psi(x)}{dx^2} - V_0 (1 - \varepsilon |x|)\psi(x) = -|E| \psi(x)$$
Change the coordinate using $x:=ay$ for some constant $a>0$.
Now you get :
$$-\frac{d^2\psi(y)}{dy^2} - a^2V_0 (1 - a\varepsilon |y|)\psi(y) = -a^2|E| \psi(y)$$
So by making $a\varepsilon=1$ (a choice we can make) :
$$-\psi''(y) - a^2V_0 (1 - |y|)\psi(y) = -a^2|E| \psi(y)$$
And you get your triangular problem back with scaled energy and potential values.
It is always a triangular potential problem.
A: 
The problem is such that the parametrized equation has the solutions of the form $\operatorname{Ai}(\varepsilon^{-2/3}(\varepsilon|x|-(1-|E|)))$, which do not converge to the usual trigonometric function but to the "decaying" one since the Airy functions converge to zero for big negative arguments. Where is the error?

You've forgotten that Schrödinger's equation is a second-order equation, and that to satisfy your boundary condition $\psi(0)=0$ you need both kinds of Airy functions: $\operatorname{Ai}$ and $\operatorname{Bi}$. The solution of your equation for the potential $-V_0(1-\varepsilon x)$, satisfying the boundary condition, is then:
$$\psi(x)=\operatorname{Ai}\left(\frac{V_0(\varepsilon |x|-1)+|E|}{(V_0\varepsilon)^{2/3}}\right)-\operatorname{Bi}\left(\frac{V_0(\varepsilon |x|-1)+|E|}{(V_0\varepsilon)^{2/3}}\right)\frac{\operatorname{Ai}\left(\frac{|E|-V_0}{(V_0\varepsilon)^{2/3}}\right)}{\operatorname{Bi}\left(\frac{|E|-V_0}{(V_0\varepsilon)^{2/3}}\right)}.$$
With this expression you'll get
$$\frac{\psi(x)}{\psi'(0)}\to\frac{\sin(\sqrt{V_0-|E|}x)}{\sqrt{V_0-|E|}}\quad\text{as}\quad\varepsilon\to0.$$
