I'm in trouble to show the following:
Let $\phi(s)\in C^\infty(0,\infty)$ such that $0\le \phi(s)\le 1$ and $\phi(s)=1 (s\ge 1)$, $\phi(s)=0(s\le 1/2)$. Let $x = (x_1, x_2)$, $\psi(x)\in C_c^\infty(\mathbb R^2)$ and $\psi(x_1, 0)=0$. Then I want to show that for any $u\in L^p((\mathbb R \times (0,\infty))$, $$ \int_0^\infty \int_{\mathbb R} \frac{\partial\phi(x_2/\epsilon)}{\partial x_2} \psi(x) u(x) dx_1 dx_2 \to 0 \quad (\epsilon \to 0). $$ Here $C_c^\infty$ means $C^\infty$ class with compact support.
First notice that the pointwise limit of the integrand is $0$. Indeed, fix $x \in \mathbb{R} \times (0,\infty)$ and let $\epsilon$ be so small that $x_2/\epsilon > 1$. Now recall that since $\phi$ is constant for $s \ge 1$ its derivative is $0$ for $s > 1$.
What we are left to prove is that we can pass the limit inside the integral signs. By Lebesgue's dominated convergence Theorem, it is enough to show that the integrand is dominated by an integrable function.
This is easy: by assumption, $\phi'$ is continuous and $0$ outside the compact set $[1/2,1]$, hence it is bounded. Notice also that $u \in L^1(\text{supp}(\psi))$. Then we have $$|\phi'(x_2/\epsilon)\psi(x)u(x)| \le \big(\|\phi'\|_{L^\infty}\|\psi\|_{L^\infty}\big)\chi_{\text{supp}(\psi)}(x)u(x) \in L^1(\mathbb{R} \times (0,\infty)).$$