= Solution
Take fixed $E>0$ and define $k=\sqrt{2mE}/\hbar$. For sufficiently large $V_0$, the barrier region has $\kappa=\sqrt{2m(V_0-E)}/\hbar$ and solves $\psi''=\kappa^2\psi$. Solving this constant-<coefficient> equation gives the exact transfer relation
$$
\begin{pmatrix}\psi(a)\\\psi'(a)\end{pmatrix}
=\begin{pmatrix}\cosh(\kappa a)&\sinh(\kappa a)/\kappa\\
\kappa\sinh(\kappa a)&\cosh(\kappa a)\end{pmatrix}
\begin{pmatrix}\psi(0)\\\psi'(0)\end{pmatrix}.
$$
In the specified limit, $a=U/V_0$ and $\kappa a\to0$, while
$$
\kappa^2a\to\frac{2mU}{\hbar^2}=:\eta.
$$
The transfer <matrix> therefore tends to $\begin{pmatrix}1&0\\\eta&1\end{pmatrix}$. As the two boundaries collapse to the origin, this proves <continuity> and the <derivative> jump for a <delta potential>:
$$
\psi(0+)=\psi(0-),\qquad
\psi'(0+)-\psi'(0-)=\eta\psi(0).
$$
The vanishing exterior phase shift $e^{ika}\to1$ introduces no further factor.
Write the incoming and reflected waves as $e^{ikx}+r e^{-ikx}$ on the left and the transmitted wave as $t e^{ikx}$ on the right. Matching gives
$$
t=1+r,\qquad ikt-ik(1-r)=\eta t,
\qquad t=\frac{2ik}{2ik-\eta}=\frac1{1+i\eta/(2k)}.
$$
The denominator is nonzero for real $\eta$ and $k>0$, so the limiting matching system is well defined. The <probability currents> on either side have the same factor $\hbar k/m$, making the transmission probability $|t|^2$. Thus <scattering by a delta potential> gives
$$
\boxed{T=\frac1{1+\eta^2/(4k^2)}
=\left(1+\frac{mU^2}{2E\hbar^2}\right)^{-1}.}
$$
The limit retains a finite reflection probability despite the barrier's vanishing width because its integrated strength remains finite.
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