Thus, for each leaf $L=\{(z,r(z))\in {\Bbb C}^2:\,z\in {\Bbb
C},\,|z|>M\}\subset B$, there exist two leaves $L_1=\{(z,r_1(z))\in
{\Bbb C}^2:\,z\in {\Bbb C},\,|z|>M_1\}\subset B_1$ and $L_2=\{(z,r_2(z))\in {\Bbb C}^2:\,z\in {\Bbb C},\,|z|>M_2\}\subset
B_2$ such that $\lim_{z\rightarrow\infty}(r(z)-r_1(z))=0$ and $\lim_{z\rightarrow\infty}(r(z)-r_2(z))=0$. Then $$|r_1(z)-r_2(z)|\leq |r_1(z)-r(z)|+|r(z)-r_2(z)|_{\overrightarrow{z\rightarrow
\infty}}0.$$ Therefore, ${\overline{{\cal C}}}_1$ and ${\overline{{\cal C}}}_2$ approach each other.$\Box$

In the following, we illustrate the above results with an example.

Let $\cal C$ and $\overline{{\cal C}}$ be two plane curves defined implicitly by the polynomials $$f(x,y)=2y^3x-y^4+2y^2x-y^3-2x^3+x^2y+3\in {\Bbb R}[x,y],\qquad \mbox{and}$$ $$\overline{f}(x,y)=y^3x-y^4+y^2x-y^3-x^3+x^2y+2\in {\Bbb R}[x,y],$$ respectively. Let us prove that $\cal C$ and $\overline{{\cal C}}$ approach each other (see Figure \[Ej-approaching curves\]) at the infinity branch associated to the infinity point $P=(1:0:0)$ (note that both curves have $P$ as an infinity point). Reasoning as in Example \[Ej-ramas infinitas\], we get that the infinity branch of $\cal C$ associated to $P$ is given by $B=L_1\cup L_2\cup L_3$, where $L_i=\{(z,r_i(z))\in {\Bbb
C}^2:\,z\in {\Bbb C},\,|z|>M\}$, $$r_i(z)=c_i^2z^{2/3}-1/3+1/9c_i^2z^{-2/3}-2/81c_i^4z^{-4/3}-1/2c_i^7z^{-7/3}+\cdots$$ and $c_i,\,i=1,2,3$ are the complex roots of $x^3=1$. On the other hand, the infinity branch of $\overline{{\cal C}}$ associated to $P$ is given by $\overline{B}=\overline{L}_1\cup \overline{L}_2\cup
\overline{L}_3$, where $\overline{L}_i=\{(z,\overline{r}_i(z))\in
{\Bbb C}^2:\,z\in {\Bbb C},\,|z|>M\}$, $$\overline{r}_i(z)=c_i^2z^{2/3}-1/3+1/9c_i^2z^{-2/3}-2/81c_i^4z^{-4/3}-2/3c_i^7z^{-7/3}+\cdots$$ and $c_i,\,i=1,2,3$ are the complex roots of $x^3=1$ (to compute $r_i$ and $\overline{r}_i$, we use the [*algcurves*]{} package included in 