Consider $$c(t)=(a\cos\ (t),b\sin\ (t)),\ 0\leq t,\ a=1 > b>0$$ in $\mathbb{R}^2$
Fix $0<t_1<\pi/2$ so that $x_1=c(t_1)$
When $x_i=c(t_i)$ and $x_i$ is a positive multiple of $c'(t_{i-1})$, then how many are there limit points for subsequences in $\{x_i\}$ ?
Proof : Note that $x_{4k+i}$ are in $i$-th quadrant $\bigg\{ \frac{\pi}{2}(i-1)< \theta < \frac{\pi}{2}i\bigg\}$
When $y_n$ is a convergent subsequence in $\{x_{4k+1}\}$, then $\{ y_{n+i}\},\ 1\leq i\leq 4$ are also convergent. Here $\{y_n\},\ \{y_{n+4}\}$ share same limit ?
That is, $x_n$ has exactly $4$ limit points ?
$c'(t_1)=(-\sin\ t_1,b\cos\ t_1)$ where $0<t_1< \frac{\pi}{2}$. When $\pi/2<t_2<\pi$ s.t. $$ c(t_2) =C c'(t_1),\ C>0$$ then $$ (\cos\ t_2,b\sin\ t_2)= C(-\sin\ t_1 ,b\cos\ t_1),\ C>0 $$
so that $$ C= \frac{\cos\ t_2}{-\sin\ t_1} = \frac{\sin\ t_2}{\cos\ t_1} $$
Here $t_2=\frac{\pi}{2}+t_1$. That is, $\{x_n\}$ is a set of four points.