Limit of a sequence of unit vectors Suppose that $x_n=(a_n,b_n)$ is a sequence in $\mathbb{R}^2$ with $\lim x_n= \infty$. Let $y_1=(c_1,d_1)$ and $y_2=(c_2,d_2)$ such that:
\begin{align*}
\lim \frac{x_n-y_1}{||x_n-y_1||}=t \quad \text{and} \quad \lim \frac{x_n-y_2}{||x_n-y_2||}=s.
\end{align*}
Then, how can one show that the unit vectors $t$ and $s$ are equal? Does the difference of the normalized sequences tend to zero?
EDIT: By transfering the problem to complex numbers, it is easy to show it: just divide the sequences (assuming that $x_n \neq y_1$, $y_2$ for large values of $n$). But this is not possible when dealing with the case of $\mathbb{R}^2$, let alone $\mathbb{R}^d$. 
 A: Directly calculate $\frac{x_n-y_1}{||x_n-y_1||}-\frac{x_n-y_2}{||x_n-y_2||}$. You may observe that it suffices to show $\frac{||x_n-y_2||-||x_n-y_1||}{||x_n-y_2||||x_n-y_1||}x_n$ tends to $0$. Applying a multiplication by $||x_n-y_2||+||x_n-y_1||$ to both numerator and denominator, you can see that it is sufficient to analyse $\frac{(||x_n-y_1||^2-||x_n-y_2||^2)|x_n|}{(||x_n-y_1||+||x_n-y_2||)||x_n-y_1||||x_n-y_2||}$. Directly writting out the coordinates, you can find that $||x_n-y_1||^2-||x_n-y_2||^2$ is of at most $O(|x_n|)$. It can clearly generalize to $\mathbb{R}^n$. But I don't believe it is true for the infinite dimensional space. I cannot give a counter example now.
A: Hint: In the first limit, divide numerator and denominator by $\|x_n\|$ to get
$$
\lim \frac{\frac{x_n}{\|x_n\|} - \frac{y_1}{\|x_n\|}} {\|\frac{x_n}{\|x_n\|} - \frac{y_1}{\|x_n\|}\|}
$$
The limit of the quotients is the quotient of the limits (if the denominator limit is nonzero). The numerator limit is the difference of two terms, the second being $L = \lim y_1 / \|x_n\|$. Let's look at the norm of this limit: 
\begin{align}
\| L \| 
&= \| \lim \frac{y_1}{\|x_n\|} \| \\
&= \lim \| \frac{y_1}{\|x_n\|} \| \\
&= \lim \frac{\| y_1\|}{\|x_n\|}  \\
&= \| y_1\| \cdot 0  \\
\end{align}
because $\lim \|x_n\| = \infty$. Hence $\|L\| = 0$, so $L = 0$. 
The overall limit is therefore
\begin{align}
C_1 &= \lim \frac{x_n } {\|x_n\ - y_1\|}
\end{align}
Now letting $\theta_n$ be the angle of the vector $x_n$ (which is well defined for all sufficiently large $n$, since $x_n \to \infty$ implies that $\|x_n\| \to \infty$, hence for large $n$, we know $x_n \ne 0$) between, say, $0$ and $2\pi$. 
Then the angle of $C_1$ must be the limit of the angles of the $x_n$s, i.e., 
$$
\theta_C = \lim \frac{\theta_n}{\| x_n - y \|}.
$$
And since $C$ is a unit vector, this uniquely determines $C$: it's $(\cos t, \sin t)$, where $t = lim \theta_n$. 
Noting that this definition of $C$ is independent of $y_1$ (which didn't enter into the computation anywhere), we see that the $C$ derived from the $t$-limit and the $C$ derived from the $s$-limit must be equal, and in fact that 
$$
C = \lim \frac{x_n}{\|x_n\|}.
$$
Whew. That's kind of a mess, eh? 
Note: if the limit $\theta_C$ turns out to be $0$, then we can't guarantee continuity passing through the limit, and we have to redo everything measuring angles from the negative $x$-axis instead ... 
but that's just a repetition of the same argument, so I'm gonna skip it. 
A: Let's give an upper estimate for (the norm of)
$$ \frac{x}{||x||} - \frac{x+h}{||x+h||}$$
The expression equals:
$$\frac{ ||x|| h + (||x||- ||x+h||) x}{||x||\cdot ||x+h||}
$$
so its norm is at most 
$$\frac{ 2||h||}{||x+h||} \le \frac{2 ||h||}{||x||- ||h||}= \frac{ 2 ||h||/||x||}{1 - ||h||/||x||}$$
That should be enough for the proof. 
A: Let  $\frac {x_n-y_2}{\|x_n-y_2\|}=s+e_n=$ and $\frac {x_n-y_1}{\|x_n-y_1\|}=t+d_n$. By direct calculation, $$(\bullet)\quad s+e_n=\frac {\|x_n-y_1\|}{\|x_n-y_2\|}(t+d_n)+\frac {y_1-y_2}{\|x_n-y_2\|}.$$
Now $\|x_n-y_1\|=\|x_n-y_2+(y_2-y_1)\|$ is bounded above and below by $\|x_n-y_2\|\pm \|y_2-y_1\|$. So $\frac {\|x_n-y_1\|}{\|x_n-y_2\|}$ is bounded above and below by $1\pm \frac {\|y_1-y_2\|}{\|x_n-y_2\|}.$ And $\|x_n-y_2\|\to  \infty $ because $\|x_n-y_2\|\geq \|x_n\|-\|y_2\|.$ 
So $\frac {\|x_n-y_1\|}{\|x_n-y_2\|}\to 1$ and $\frac {y_1-y_2}{\|x_n-y_2\|}\to 0.$ And $e_n\to 0$ and $d_n\to 0$. Applying these to $(\bullet)$ , we have $s=t$.
