Let $a_n$ and $b_n$ be sequences of real numbers. If $b_n$ is bounded and $\lim_{n \to \infty} a_n = 0$, then $\lim_{n\to\infty} a_{n}b_{n} = 0$ Prove the below statement:
Let $a_n$ and $b_n$ be sequences of real numbers. If $b_n$ is bounded and $\lim_{n \to \infty} a_n = 0$, then $\lim_{n \to \infty} a_n b_n=0$
When I read this question, I read that $b_n$ may or may not converge, so taking the example $\dfrac{1}{n}$ and $3n$, $\lim_{n\to\infty} a_nb_n=3\neq0$. What am I getting wrong? It's a theorem from my book and the only hint is that $b_n$ being bounded is crucial. So I know it must be true but I can't understand why. 
Any suggestions or further hints in the right direction would be greatly appreciated. This is the last problem I have and have been stumped by it all day. 
 A: Since $b_n$ is bounded, there is a number $M$ such that for all $n$, $|b_n|\leq M$. Now 
$$0 \leq |a_n b_n|\leq |a_n| M.$$
Take the limit as $n\to \infty$ and you are done.
A: $|a_{n}b_{n}-ab|=|a_{n}b_{n}-ab_{n}+ab_{n}-ab|\leq |a_{n}b_{n}-ab_{n}| + |ab_{n}-ab|$
$=|b_{n}||a_{n}-a|+|a||b_{n}-b|=|b_{n}||a_{n}-a| + 0|b_{n}-b|=|b_{n}||a_{n}-a|$
The sequence $b_{n}$ is bounded i.e.
$|b_{n}|<B<\infty$
then 
$|a_{n}b_{n}-ab|<B|a_{n}-a|$
$a_{n}$ is a convergent sequence so $\forall \varepsilon'>0$ $\exists N \in \mathbb{N}$ such that $\forall n \geq N$ $|a_{n}-a|<\varepsilon'$. Choose $\varepsilon'=\frac{\varepsilon}{B}$, then 
$|a_{n}b_{n}-ab|<\varepsilon$ for all $n\geq N$
$a$ is zero and so 
$|a_{n}b_{n}-0|<\varepsilon$ for all $n\geq N$
and so $a_{n}b_{n}\rightarrow 0$ as $n\rightarrow \infty$
A: What does it means that $$\lim_{n\to\infty}a_n=0\,\text{?}$$
It means that for any arbitrarily small real $\varepsilon>0$, after some $n_\varepsilon\in\mathbb N$, all $|a_n|$ are less than $\varepsilon$, for $n>n_\varepsilon$.
What does it means the $\{b_n\}$ is bounded?
It means that there is a boundary $B\in\mathbb R$ so that $|b_n|<B$.
Hint:
For a given $\varepsilon>0$, take $n_{\varepsilon/B}$.
A: Suppose the upper bound of $b_n$ is the constant $c$. Now suppose that $b_n = c$.
$$\lim_{n\to\infty} ca_n=c\lim_{n\to\infty} a_n=0$$
