approach each other at the infinity.

The results presented in this section will be used in Section 5, where a method to compare the asymptotic behavior of two curves is developed.

Given two leaves, $L=\{(z,r(z))\in
{\Bbb C}^2:\,z\in {\Bbb C},\,|z|>M\}$ and $\overline{L}=\{(z,\overline{r}(z))\in {\Bbb C}^2:\,z\in {\Bbb
C},\,|z|>\overline{M}\}$, we say that they are convergent if $\lim_{z\rightarrow\infty} (\overline{r}(z)-r(z))=0.$

\[L-DistVertical\] Two leaves $L=\{(z,r(z))\in {\Bbb C}^2:\,z\in {\Bbb C},\,|z|>M\}$ and $\overline{L}=\{(z,\overline{r}(z))\in {\Bbb C}^2:\,z\in {\Bbb
C},\,|z|>\overline{M}\}$ are convergent if and only if the terms with non negative exponent in the series $r(z)$ and $\overline{r}(z)$ are the same.

**Proof:** Let $$r(z)=mz+a_1z^{\frac{N-N_1}{N}}+a_2z^{\frac{N-N_2}{N}}+\cdots ,\,\, N, N_i\in {\Bbb N},\, 0<N_1<N_2<\cdots,\, \, a_i\not=0$$ and $$\overline{r}(z)=\overline{m}z+b_1z^{\frac{\overline{N}-\overline{N}_1}{\overline{N}}}+b_2z^{\frac{\overline{N}-\overline{N}_2}{\overline{N}}}
+\cdots,\,\, \overline{N}, \overline{N}_i\in {\Bbb N},\,
0<\overline{N}_1<\overline{N}_2<\cdots,\,\,b_i\not=0.$$ Then, $$r(z)-\overline{r}(z)=mz-\overline{m}z+a_1z^{\frac{N-N_1}{N}}-b_1z^{\frac{\overline{N}-\overline{N}_1}{\overline{N}}}
+a_2z^{\frac{N-N_2}{N}}-b_2z^{\frac{\overline{N}-\overline{N}_2}{\overline{N}}}+\cdots.$$ Note that $\lim_{z\rightarrow\infty}(r(z)-\overline{r}(z))=0$ if and only if $r(z)-\overline{r}(z)$ has no terms with non negative exponent. This situation holds if the terms with non negative exponent in both series, $r(z)$ and $\overline{r}(z)$, are the same. $\Box$

\[R-DistVertical\]

1.  From Lemma \[L-DistVertical\], we deduce that $m=\overline{m}$ and then, $L$ and $\overline{L}$ are associated to the same infinity point.

2.  Note that the number of terms with positive exponent in both series is finite.

\[D-distance0\] Two infinity branches, $B$ and $\overline{B}$, are convergent if there exist two convergent leaves $L\subset B$ and 