Using the definition, that for every $\epsilon>0$, there exists a $N\in\mathbb{N}$, so that $d(v_i,L)$ for all $i\geq N$, where $v_i$ is the $i$th vector in the sequence. And $L$ is the limit.
I assumed that let there be two limits, $L$ and $M$.
Now I choose my epsilon to be $\frac13d(L,M)$, but after that, how do I proceed? Plain algebra doesn't help.
Note that $d(a,b)$ is the distance between vectors $a$ and $b$ in $\mathbb{R^n}$.
I like to think about it geometrically first. Your sequence points are getting really close to $L$ and to $M$ at the same time, but these two points are a positive distance away. Eventually, they're going to have to choose a point to get closer to.
In fact, if you choose $\varepsilon = d(L, M)/2$, then the open ball centred at $L$ with radius $\varepsilon$ doesn't even intersect the open ball centred at $M$ with radius $\varepsilon$. This fact comes courtesy of the triangle inequality. See if you can prove this by contradiction. Then, see if you can apply the limit definition to show why this a problem for this sequence.
