According to Proposition 3.2 we can choose $s$ big enough so that the $\mathrm{Log}_{s}$-image of every hypersurface $H_{s}^{v}(a)$ lies in the $\epsilon$-neighborhood of $U^{\epsilon}_{v}$. Recall from Lemma 3.3 that a small neighborhood of $U^{\epsilon}_{v}$ lies in the domain $Q^{\lambda}_{(\{0\}\mid v)}(\epsilon)$. Thus we can assume that all hypersurfaces $H_{s}^{v}(a)$ lie entirely in $\mathrm{Log}_{s}^{-1}(Q^{\lambda}_{(\{0\}\mid v)}(\epsilon))$.

Whenever $\mathrm{Log}_{s}(x)\in Q^{\lambda}_{(\{0\}\mid v)}(\epsilon)$, we have

$$\langle m,\frac{\log|x|}{\log|s|}\rangle+\lambda(m)\leq\lambda(0)-\epsilon,\text{ for all }m\neq v,\{0\}$$

or, equivalently,

$$|x^{m}s^{\lambda}(m)|\leq|s|^{-\epsilon}|s|^{\lambda(0)}.$$

This means that the values of all monomials $x^{m}s^{\lambda(m)},\ m\neq v,\{0\}$, for $x\in\mathrm{Log}_{s}^{-1}(Q^{\lambda}_{(\{0\}\mid v)}(\epsilon))$, are (uniformly) bounded by $|s|^{-\epsilon}|s|^{\lambda(0)}$. Note also, that their log-derivatives are bounded by $C|s|^{-\epsilon}|s|^{\lambda(0)}$, some constant $C\geq 0$, since

$$x\frac{\partial}{\partial x}(a_{m}s^{\lambda(m)}x^{m})=m\cdot a_{m}s^{\lambda(m)}x^{m}.$$

For any basis $\{e_{i}\}$ of $(\mathbb{Z}^{d})^{*}$, the functions $y_{i}=x^{e_{i}}$ give affine coordinates on $(\mathbb{C}\backslash\{0\})^{d}$. We choose $e_{1}=-v$, multiply the equations of the hypersurfaces in our family $H_{s}^{v}(a)$ by $y_{1}=x^{-v}$, and look for critical points:

$$\frac{\partial}{\partial y_{1}}\bigl(y_{1}s^{\lambda(0)}+s^{\lambda(v)}+y_{1}\sum\limits_{m\neq\{0\},v}a_{m}s^{\lambda(m)}x^{m}\bigr)\\ =s^{\lambda(0)}+\bigl(1+y_{1}\frac{\partial}{\partial y_{1}}\bigr)\sum\limits_{m\neq\{0\},v}a_{m}s^{\lambda(m)}x^{m}=s^{\lambda(0)}(1+O(|s|^{-\epsilon}))\neq 0,$$

for large enough $s$. Thus, there are no critical points, hence every member of our family $H_{s}^{v}(a)$ is smooth.

Finally, note that $\bigcup_{q\in U^{\epsilon}_{v}}\mathcal{F}_{q}$ is in $Q_{v}^{\lambda^{\epsilon}}$, but Lemma 3.6 asserts that the vectors $\xi\in\mathfrak{X}$ in $Q_{v}^{\lambda^{\epsilon}}$ satisfy $\langle v,\xi\rangle=1$. Thus, for any point of intersection $H_{s}^{v}(a)\cap X_{s}(q)$ the corresponding tangent vector to $X_{s}(q)$ has the form:

$$\bar{\xi}=y_{1}\frac{\partial}{\partial y_{1}}+\alpha_{2}y_{2}\frac{\partial}{\partial y_{2}}+\dots+\alpha_{d}y_{d}\frac{\partial}{\partial y_{d}}.$$

Differentiating the defining equation for $H_{s}^{v}(a)$ with respect to $\bar{\xi}$ gives:

$$\bar{\xi}\bigl(y_{1}s^{\lambda(0)}+s^{\lambda(v)}+y_{1}\sum\limits_{m\neq\{0\},v}a_{m}s^{\lambda(m)}x^{m}\bigr)\\ =y_{1}s^{\lambda(0)}(1+O(|s|^{-\epsilon}))+\sum\limits_{i=2}^{d}\alpha_{i}y_{i}\frac{\partial}{\partial y_{i}}\bigl(y_{1}\sum\limits_{m\neq\{0\},v}a_{m}s^{\lambda(m)}x^{m}\bigr)\\ =y_{1}s^{\lambda(0)}(1+O(|s|^{-\epsilon}))+\sum\limits_{i=2}^{d}y_{1}s^{\lambda(0)}O(|s|^{-\epsilon})=y_{1}s^{\lambda(0)}(1+O(|s|^{-\epsilon}))\neq 0.$$

Thus, we can conclude that $\bar{\xi}$ is transversal to the tangent planes to $H_{s}^{v}(a)$, that is $X_{s}(q)$ is transversal to all $H_{s}^{v}(a)$. ∎