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.
 A: HINT: If $x_n$ is bounded, then $\exists M>0: |x_n|<M$ for all $n$. Hence $|x_ns_n|<M\varepsilon$ for large $n$. It should be clear that $x_ns_n$ also tends to 0.
A: Since ${x_n}$ is a bounded sequence, then for every $N_1\in \mathbb N$  there is M >0 $\in \mathbb R$ such that when n> $N_1$,$|x_n|$< M.We also know that since $lim_{n\rightarrow\infty} s_n$ =$0$, for every $\epsilon$ >0 $\in \mathbb R$,there is an $N_2\in \mathbb N$ such that when n> $N_2$, $|s_n|$< $\epsilon$. Let $\epsilon$'= M$\epsilon$ and choose n > max {$N_1$,$N_2$}. By the Cauchy Schwarz inequality:
$|x_n s_n|\leq$ $|x_n||s_n|$ < M$\epsilon$ = $\epsilon$'.
So $lim_{n\rightarrow\infty} (x_n s_n)$ = 0.
That does it.
Addendum: Andre Nicolas correctly pointed out that ($x_n$) need not itself have a limit even though it's bounded. While this is very true, the argument above is completely general and refers to the values of the product of the sequence ranges. I've clarified some steps above to show this. 
A: 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.
A: We write a standard $\epsilon$-$N$ proof. One could alternately use a squeezing argument.
Since the sequence $(x_n)$ is bounded, there is a $B\gt 0$ such that $|x_n|\le B$ for all $n$. It follows that
$$|s_nx_n-0|\le B|s_n|\tag{1}$$
 for all $n$.
We want to show that given $\epsilon\gt 0$, there is an $N$ such that $|s_nx_n-0|\lt \epsilon$ for all $n\gt N$. By Inequality (1), it is enough to show that there is an $N$ such that if $n\gt N$ then $B|s_n|\lt \epsilon$.
Since the sequence $(s_n)$ has limit $0$, there is an $N$ such that if $n\gt N$ then $|s_n|\lt \frac{\epsilon}{B}$. It follows that if $n\gt N$ then $B|s_n|\lt B\cdot\frac{\epsilon}{B}=\epsilon$, and therefore $|s_nx_n-0|\lt \epsilon$.
