Lemma 5.3 and the inequality (4) imply that $\mathcal{A}_{t}$ converge to a subset of $\mathcal{A}_{K}$. Indeed, for each $t$ we can rewrite $|a_{j}t^{v(j)}z^{j}|$ as $|t^{c_{j}}z^{j}|$, $c_{j}=v(j)+\log_{t}|a_{j}|$. Such a monomial induces a linear function $c_{j}+jx$ in $\mathbb{R}^{n+1}$. The inequalities

where $N+1$ is the number of monomials in $f_{t}$, cut out a uniformly bounded neighborhood of $\mathcal{A}_{K}$ which contains $\mathcal{A}_{K}$.

The limit of $\mathcal{A}_{t}$ cannot be any smaller than $\mathcal{A}_{K}$ by the following topological reason. A component of the complement of the set described by the inequalities (5) is given by the inequality $c_{k}+kx>\max\limits_{j\neq k}(c_{j}+jx)+\log_{t}(N)$. By [3] this component is contained in the component of $\mathbb{R}^{n+1}\smallsetminus\mathcal{A}_{t}$ corresponding to the index $k$. Thus, different components of the set described by (5) must be contained in different components of $\mathbb{R}^{n+1}\smallsetminus\mathcal{A}_{t}$. ∎