I believe this is really simple, but I can't figure it out alone. Let $\{x_k\}_{k\in \mathbb{N}}\subset \mathbb{R}^n$ and be a sequence such that $\|x_k\|\to \infty$, and let $v\in \mathbb{R}^n$ be arbitrary.
Is the set of limit points of $\left\{ \frac{x_k}{\left\|v+x_k\right\|} \right\}_{k\in \mathbb{N}}$ equal to the set of limit points of $\left\{ \frac{x_k}{\left\|x_k\right\|} \right\}_{k\in \mathbb{N}}$?
It looks true, because $v$ is a finite perturbation and everything tells me that $x_k$ is going to lead the convergence since it diverges, but I couldn't give formal arguments for proving that. I appreciate any help.
If you wish to show two sequences $(a_n)$ and $(b_n)$ have the same limit points, then it suffices to show that $\|a_n - b_n\| \to 0$ as $n \to \infty$. If we have a limit point $L$ of $(a_n)$, then we have a subsequence $a_{n_k} \to L$. Then, $$\|b_{n_k} - L\| \le \|b_{n_k} - a_{n_k}\| + \|a_{n_k} - L\| \to 0 + 0,$$ since $\|b_{n_k} - a_{n_k}\|$ is a subsequence of the convergent sequence $\|a_n - b_n\|$ and $\|a_{n_k} - L\| \to 0$. It follows that $b_{n_k} \to L$. So, every limit point of $(a_n)$ is a limit point of $(b_n)$. Symmetrically, we can reverse the roles of $(a_n)$ and $(b_n)$ to obtain the reverse inclusion, i.e. the limit points of $(a_n)$ and $(b_n)$ are the same.
Thus, it suffices to show that $$\left\|\frac{x_k}{\|v + x_k\|} - \frac{x_k}{\|x_k\|}\right\| \to 0.$$ Consider: \begin{align*} \left\|\frac{x_k}{\|v + x_k\|} - \frac{x_k}{\|x_k\|}\right\| &= \|x_k\|\left|\frac{1}{\|v + x_k\|} - \frac{1}{\|x_k\|}\right| \\ &= \|x_k\|\frac{\Big|\|x_k\| - \|v + x_k\|\Big|}{\|v + x_k\|\|x_k\|} \\ &\le \frac{\|v\|}{\|v + x_k\|} \\ &\le \frac{\|v\|}{\|x_k\|- \|v\|} \to 0. \end{align*}