Limits of two sequences of vectors being parallel? If ${u_k}$ is a sequence in a vector space $X$ with $u_k \to 0$, and there exist $v, w \in X$ and two real sequences $t_k \to 0, s_k \to 0$, such that $\frac{u_k}{t_k} \to v$ and $\frac{u_k}{s_k} \to w$, must it be the case that $v$ and $w$ are colinear, i.e., $\exists \lambda \in \mathbb{R}$, s.t. $ v = \lambda w$?
I'm having trouble proving this, but also can't come up with counter-examples...
P.S. By $ x_k \to y$, of course I mean $\lim_{k\to\infty}x_k = y$. And I'm currently looking just at $X$ being the Euclidean space with the standard $\ell_2$ norm/metric.
 A: Let $(s_k)_{k\in\Bbb{N}}$ and $(t_k)_{k\in\Bbb{N}}$ be two nonzero real sequences and $(u_k)_{k\in\Bbb{N}}$ a sequence in $X$ such that
$$\lim_{k\to\infty}s_k=\lim_{k\to\infty}t_k=0,
\qquad
\lim_{k\to\infty}u_k=0,
\qquad
\lim_{k\to\infty}\tfrac{u_k}{s_k}=v
\qquad\text{ and }\qquad
\lim_{k\to\infty}\tfrac{u_k}{t_k}=w.$$
If either $v=0$ or $w=0$ then $v$ and $w$ are collinear. So suppose that $v$ and $w$ are nonzero.
Because the inner product on $X$ is continuous we have
\begin{eqnarray*}
||v||^2=\langle v,v\rangle
&=&\lim_{k\to\infty}\langle\tfrac{u_k}{s_k},\tfrac{u_k}{s_k}\rangle
=\lim_{k\to\infty}\tfrac{||u_k||^2}{s_k^2}\\
\langle v,w\rangle
&=&\lim_{k\to\infty}\langle\tfrac{u_k}{s_k},\tfrac{u_k}{t_k}\rangle
=\lim_{k\to\infty}\tfrac{||u_k||^2}{s_kt_k},
\end{eqnarray*}
where $\langle v,v\rangle\neq0$. It follows that
$\tfrac{\langle v,w\rangle}{||v||^2}
=\lim_{k\to\infty}\tfrac{s_k}{t_k}$, and hence that
$$\frac{\langle v,w\rangle}{||v||^2}v
=\lim_{k\to\infty}\tfrac{s_k}{t_k}\lim_{k\to\infty}\tfrac{u_k}{s_k}
=\lim_{k\to\infty}\tfrac{u_k}{t_k}=w,$$
which shows that $w$ is a scalar multiple of $v$.
A: This indeed holds. Let $(t_k)_{k\in \mathbb{N}}$ be a non-zero sequence, $t_k >0$ for all $k$ (the case of a negative sequence is similar), $(u_k)_{k\in \mathbb{N}} \in X, \lim_{k\to\infty} u_k = 0$, and $\lim_{k\to\infty} \frac{u_k}{t_k}=v\neq 0$ (the statement trivially holds if $v=0$).
We just need one simple lemma, a limit law for a real and a vector sequence (this was used implicitly in @Servaes's answer):  Let $(a_k)_{k\in \mathbb{N}}$ be a sequence of reals, $(b_k)_{k\in \mathbb{N}}$ be a sequence in a vector space $X$, $a \in \mathbb{R}$, $b \in X$, such that $\lim_{k\to\infty} 
 a_k = a, \lim_{k\to\infty} b_k = b$. Then $\lim_{k\to\infty} a_k b_k = ab$.
Now observe that $\lim_{k\to\infty} \frac{u_k}{t_k}=v$ implies $\lim_{k\to\infty} \frac{\|u_k\|}{t_k} = \lim_{k\to\infty} \frac{\|u_k\|}{|t_k|}=\|v\|$, so $\lim_{k\to\infty} \frac{t_k}{\|u_k\|} = \frac{1}{\|v\|}$ (since $v\neq 0$). By our lemma, we then have $$\lim_{k\to\infty} \frac{u_k}{\|u_k\|} =  \lim_{k\to\infty} \frac{u_k}{t_k} \frac{t_k} {\|u_k\|} = (\lim_{k\to\infty} \frac{u_k}{t_k}) (\lim_{k\to\infty} \frac{t_k} {\|u_k\|}) = \frac{v}{\|v\|} $$
If we also have $\lim_{k\to\infty} \frac{u_k}{s_k}=w\neq 0$ (again assuming $s_k > 0$; you need a negative sign in the following limit if $s_k < 0$), then by the same argument as above,
$$\lim_{k\to\infty} \frac{u_k}{\|u_k\|} = \frac{w}{\|w\|} $$
By the uniqueness of limits, we must have 
$$ \frac{v}{\|v\|} = \frac{w}{\|w\|}$$
(or, $\frac{v}{\|v\|} = - \frac{w}{\|w\|}$, when $t_k$ and $s_k$ have different signs), which shows that $u$ and $w$ are parallel (and in fact point in the same direction as $\lim_{k\to\infty} \frac{u_k}{\|u_k\|}$).
