= Solution
Put $p=\sqrt{2mE}/\hbar$ and $\gamma=mk/\hbar^2$. Integrating the <Schrödinger equation> across the <delta potential> gives continuity of $\psi$ and
$$
\psi'(0+)-\psi'(0-)=2\gamma\psi(0).
$$
For incidence from the left, write $\psi=e^{ipx}+r e^{-ipx}$ on $x<0$ and $\psi=t e^{ipx}$ on $x>0$. Matching gives $t=1+r$ and $ip(t-1+r)=2\gamma t$, hence
$$
r=-\frac{i\gamma}{p+i\gamma},\qquad t=\frac p{p+i\gamma}.
$$
The incident and transmitted <probability currents> have the same velocity factor, so the <reflection coefficient> and <transmission coefficient> are
$$
\boxed{R=\frac{\gamma^2}{p^2+\gamma^2},\qquad T=\frac{p^2}{p^2+\gamma^2},\qquad R+T=1.}
$$
For $k<0$, an even <bound state> has $\psi=Ae^{-\kappa|x|}$, $\kappa=\sqrt{-2mE}/\hbar$. The <derivative> jump now gives $-2\kappa A=2\gamma A$, so $\kappa=-\gamma>0$ and
$$
\boxed{E=-\frac{mk^2}{2\hbar^2}.}
$$
There is exactly one positive value of $\kappa$, hence exactly one even <bound state>. Its normalized amplitude is $A=\sqrt\kappa$. An odd decaying solution would have to vanish at zero and is therefore trivial.
In the shrinking <finite square well>, its integrated potential is $k=-2Ul$, not $-Ul$: the full width is $2l$. With $Ul$ fixed and $l\to0$, $0\leq(\alpha l)^2\leq2mUl^2/\hbar^2\to0$. The matching condition then bounds $\beta$ by $(2mUl/\hbar^2)(1+o(1))$, so $E$ remains bounded. Consequently $El\to0$, and the previous matching condition gives
$$
\beta=\alpha\tan(\alpha l)\sim\alpha^2l=\frac{2m(U+E)l}{\hbar^2}\longrightarrow-\frac{mk}{\hbar^2}.
$$
Using $E=-\hbar^2\beta^2/(2m)$ recovers the same bound-state energy. All other even levels disappear as the width shrinks.
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