Help with proving if $s_n$ converges to $0$ and $x_n$ is a bounded sequence, then $\lim(x_ns_n)=0$ 
Prove: If $s_n$ converges to $0$ and $x_n$ is a bounded sequence, then $\underset{n \to \infty}{\lim}(x_ns_n)=0$

I'm have trouble getting started on this proof.  
Since I know $s_n$ converges to $0$ I feel like I should start using the epsilon definition with $x_n$ 
Since $\underset{n \to \infty}{\lim} s_n=0$
So $\left|x_n-s_n\right| \lt \epsilon \implies (=) \quad \left|x_n\right| \lt \epsilon$
But I'm not sure if the aforementioned steps are correct and where to go from here.

There is another simple way if you wish to avoid $\epsilon, \delta$ stuff. You need to use the Squeeze Theorem then.
If $x_{n}$ is bounded then there is a $K > 0$ such that $|x_{n}| < K$ for all $n$. Now we can see that $$0 \leq |x_{n}s_{n}| \leq K|s_{n}|$$ and letting $n \to \infty$ and using Squeeze Theorem we get $$\lim_{n \to \infty}|x_{n}s_{n}| = 0$$ Next we have $$-|x_{n}s_{n}| \leq x_{n}s_{n} \leq |x_{n}s_{n}|$$ and again letting $n \to \infty$ and using Squeeze Theorem we get $$\lim_{n \to \infty}x_{n}s_{n} = 0$$
BTW the proof via $\epsilon,\delta$ is much easier than the above if you understand it and is the preferred approach.
