$$\label{Rg68231opp}
{\mbox{$u=0$ in~$ (-2,2)\times(-h,0)$.}}$$ Let $E_{00}$ be the second blow-up of $E_u$. Then, $$\begin{aligned}
\label{FAG:1}
&& {\mbox{if~$E_{00}={\ensuremath{\mathbb{R}}}\times(-\infty,0)\times(-\infty,0)$, then }}
\lim_{x_2\searrow0} u(0,x_2)<0;
\\
\label{FAG:2}\nonumber
&& {\mbox{if~$E_{00}=\big({\ensuremath{\mathbb{R}}}\times(-\infty,0)\times(-\infty,0)\big)
\cup \big({\ensuremath{\mathbb{R}}}\times(0,+\infty)\times{\ensuremath{\mathbb{R}}}\big)$,}}\\
&&{\mbox{then }}\lim_{x_2\searrow0} u(0,x_2)>0.\end{aligned}$$

We focus on the proof of , since the proof of  is similar. We recall (see [@2019arXiv190405393D]) that, as $k\to+\infty$, $$\label{INCLA0-QUI}
\begin{split}&{\mbox{up to a subsequence, $\chi_{k E_u}$ converges to~$\chi_{{E_{00}}}$
in~$L^1_{\rm loc}({\ensuremath{\mathbb{R}}}^3)$,}}\\
&{\mbox{and $kE_u$ converges to~$E_{00}$
locally in the Hausdorff distance.}}\end{split}$$ We claim that for every $M>0$ there exists $k_M\in{\ensuremath{\mathbb{N}}}$ such that if $k{\geqslant}k_M$ then $$\label{INCLA1}(
kE_u)\cap B_M\subseteq\left\{x_2<\frac1M\right\}.$$ To check this, we argue for a contradiction and suppose that, for some $M>1$, there are infinitely many $k$’s for which there exists $p^{(k)}=(p^{(k)}_1,p^{(k)}_2,p^{(k)}_3)\in
(kE_u)\cap B_M$ with $p^{(k)}_2{\geqslant}\frac1M$. We observe that $B_{1/(2M)}(p^{(k)})$ cannot be contained in $kE_u$, otherwise, recalling the structure of $E_{00}$ in , $$\begin{aligned}
&& \int_{ B_{1/(2M)}(p^{(k)}) }|\chi_{k E_u}(x)-\chi_{{{E_{00}}}}(x)|\,dx=
\int_{ B_{1/(2M)}(p^{(k)}) }|\chi_{k E_u}(x)|\,dx\\&&\qquad \qquad = |B_{1/(2M)}(p^{(k)})|=
|B_{1/(2M)}|,\end{aligned}$$ which is in contradiction with .

As a result, there exists $q^{(k)}\in B_{1/(2M)}( p^{(k)} )$ with $q^{(k)}\in\partial (k E_u)$. In particular, we have that $q^{(k)}_2{\geqslant}p^{(k)}_2-\frac1{2M}{\geqslant}\frac1{2M}$, whence, using the clean ball condition in [@MR2675483], there exist $c\in(0,1)$, $r_0\in\left(0,\frac1{4M}\right)$ and $\tilde q^{(k)}\in B_{r_0}( q^{(k)})$ such that $B_{cr_0}( \tilde q^{(k)})\subseteq
(kE_u)\cap B_{r_0}( q^{(k)})$.

We remark that if 