\begin{proposition}[the limit of the criterion function]\label{prop:Ninf}
Under Theorem~\ref{thm:C}, after its rotations, let $N_n=B_n+(W_n/W_n')B_n'$ be the function $N$
of Section~\ref{sec:criterion} for $(W_n,B_n)$ and $w_0=0$. Then
\[
  N_n\ \longrightarrow\ N_\infty:=\frac{(1-w)^3}{4(1+w)}\qquad\text{locally uniformly in }\mathbb D .
\]
If moreover every $(W_n,B_n)$ is a Blaschke model, then, locally uniformly in $w\in\mathbb D$,
\[
  \int_0^{2\pi}\frac{1-|w|^2}{|e^{i\theta}-w|^2}\,\log|\gamma_n(e^{i\theta})|\,\frac{d\theta}{2\pi}\ \longrightarrow\ \log|N_\infty(w)|
  =3\log|1-w|-\log|1+w|-\log4 .
\]
\end{proposition}

\begin{proof}
$W_nB_n\to\kappa\,(1-w)^2/4=w/4$ locally uniformly, so $(W_nB_n)'\to1/4$ and $W_n'\to\kappa'$ locally uniformly. Since
$\kappa'(w)=(1+w)/(1-w)^3$ has no zero in $\mathbb D$, $W_n'$ is bounded away from $0$ on each compact set for large $n$, and
$N_n=(W_nB_n)'/W_n'\to1/(4\kappa')=N_\infty$. For a Blaschke model, $(W_nB_n)'=W_n'N_n$ has no zero on the closed disk, so $N_n$ is
zero-free on a neighbourhood of it and $\log|N_n|$ is harmonic there. The Poisson formula, with $|N_n|=|B_n||\gamma_n|=|\gamma_n|$ on
the circle, shows that the left-hand side equals $\log|N_n(w)|$; and $\log|N_n|\to\log|N_\infty|$ locally uniformly, as $N_\infty$
has no zero in $\mathbb D$.
\end{proof}

\begin{remark}\label{rem:Ninf}
At $w=0$ the second statement is the gain identity (Proposition~\ref{prop:gain}), since $N_n(0)=B_n(0)\to1/4$. The limit has a pole
at the tip $w=-1$, where $\kappa'$ vanishes, while for Blaschke models every $N_n$ is bounded on the closed disk; so for them the convergence is not uniform on
$\mathbb D\cap U$ for any neighbourhood $U$ of $-1$. The function $(1-w)^3/(1+w)$ is outer, so $\log|N_\infty|$ is the Poisson integral of $-G$, where
\[
  G(s):=-\log|N_\infty(e^{is})|=\log\frac{|\cos(s/2)|}{|\sin(s/2)|^3}=\log4+\log|\kappa'(e^{is})| ,
\]
and the mean of $G$ is $\log4$. Thus the Poisson averages of $\log|\gamma_n|$ converge to those of $-G$; here $G$ is positive
for $|s|<1.944\ldots$ and tends to $-\infty$ at the tip. The measures $\log|\gamma_n(e^{i\theta})|\,d\theta$ themselves have no
weak-$*$ convergent subsequence: by Corollary~\ref{cor:cost} the means of the positive and negative parts $\ell_n^{\pm}$ of $\ell_n=-\log|\gamma_n|$ both tend
to infinity, so the
total variation tends to infinity, while a weak-$*$ convergent sequence of measures is bounded.
\end{remark}