Question: > 
> **Disprove:** If $\{x\_n\}$ is any sequence, and $\{s\_n\}$ is a sequence converging to $0$, then $$\underset{n \rightarrow \infty}{\lim} x\_n s\_n=0$$
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I'm not sure how to go about solving this proof. I tried a couple of different things, but I just realized I've been solving proofs via rote and I don't really understand the solution to a proof.

Thus, attempting to disprove a problem exposed some gaps in my understanding as I don't know what I'm trying to disprove.

First, to prove a limit problem I suppose I'm trying to show that there is some $n$ (which I visualize as being the x coordinate) that is greater than some $N$ (which I visualize as being some point on the x coordinate). $\epsilon$ just tells me how far past that point (y coordinate) I am; thus $n$ and $N$ must be less than $\epsilon$.

For this problem then I figured I just need to show that $x\_n$ is unbounded, because and unbounded $x\_n$ means the function doesn't converge to $0$. 

Is disproving this as simple as taking the product of the limits and showing that it DNE?

Answer: 
Let $x\_n = n$ and $s\_n = \displaystyle\frac{1}{n}$.

Then clearly $\underset{n \to \infty}{\lim} s\_n = 0$ as required. 

However, for every $n$, $x\_n s\_n = 1$.

Therefore $\underset{n \to \infty}{\lim} x\_n s\_n =1$.

The reason why this counterexample works is because the sequence $x\_n$ is *not* bounded. The statement is actually true for $x\_n$ bounded.
Let $x\_n =1/ s\_n$. Limit is $1$.