$B$ and $\overline{B}$ are convergent. Then, by definition, there exist two leaves $L=\{(z,r(z))\in {\Bbb
C}^2:\,z\in {\Bbb C},\,|z|>M\}\subset B$ and $\overline{L}=\{(z,\overline{r}(z))\in {\Bbb C}^2:\,z\in {\Bbb
C},\,|z|>\overline{M}\}\subset \overline{B}$ such that $\displaystyle\lim_{z\rightarrow\infty}(r(z)-\overline{r}(z))=0$. Therefore, $$\lim_{z\rightarrow\infty}d((z,r(z)),\overline{\cal C})\leq
\lim_{z\rightarrow\infty}d((z,r(z)),(z,\overline{r}(z)))=\lim_{z\rightarrow\infty}(r(z)-\overline{r}(z))=0.\qquad \mbox{\hfill $\Box$}$$

1.  From Theorem \[T-curvas-aprox\], we get that “[*proximity*]{}” is a symmetric relation; i.e., ${\overline{{\cal C}}}$ approaches ${\cal C}$ at some infinity branch $B$ iff ${\cal C}$ approaches ${\overline{{\cal C}}}$ at some infinity branch $\overline{B}$. In the following, we say that ${\cal C}$ and ${\overline{{\cal C}}}$ approach each other or that they are [*approaching curves*]{}.

2.  Theorem \[T-curvas-aprox\] and Remark \[R-common-inf-point\] imply that two approaching curves have a common infinity point.

3.  From Theorem \[T-curvas-aprox\] and Proposition \[P-convergent-branches\], we get that ${\overline{{\cal C}}}$ [*approaches*]{} ${\cal C}$ at an infinity branch $B$ if and only if for every leaf $L=\{(z,r(z))\in {\Bbb C}^2:\,z\in {\Bbb
    C},\,|z|>M\}\subset B$, it holds that $\displaystyle\lim_{z\rightarrow\infty}d((z,r(z)),\overline{\cal
    C})=0$.

\[C-approaching-curves\] Let $\cal C$ be an algebraic plane curve with an infinity branch $B$. Let ${\overline{{\cal C}}}_1$ and ${\overline{{\cal C}}}_2$ be two different curves that approach $\cal C$ at $B$. Then ${\overline{{\cal C}}}_1$ and ${\overline{{\cal C}}}_2$ approach each other.

**Proof:** From Theorem \[T-curvas-aprox\], there exist two infinity branches $B_1\subset {\overline{{\cal C}}}_1$ and $B_2\subset
{\overline{{\cal C}}}_2$, convergent with $B$. 