If $(a_n)\to A\neq 0$ and $(a_n b_n)\to AB$ then $(b_n)\to B$ 
Let $(a_n)$ and $(b_n)$ be sequences. If $(a_n) \to A\neq 0$ and $(a_n b_n)\to AB$ then $(b_n)\to B$

I know that we need to show this for both $A > 0$ and $A < 0$. But I am having problems using the assumptions to deduce that 
$$|b_n - B| < \epsilon$$
I know that since $<a_n>\to A$ then there exists an $N_1\in \mathbb{N}$ such that for all $n > N_1$
$$|a_n - A| < \epsilon$$
Similarly, since $<a_n b_n>\to AB$ then there exists an $N_2\in\mathbb{N}$ such that for all $n > N_2$
$$|a_n b_n - AB| < \epsilon$$
Any help would be appreciated.
Background information:
Theorem 8.7 - If the sequence $(a_n)$ converges to $A$ and the sequence $(b_n)$ converges to $B$ then the sequence $(a_n b _n)$ converges to $AB$
Lemma 8.1 - If $a_n\neq 0$ for all $n\in\mathbb{N}$ and if $(a_n)$ converges to $A\neq 0$ then the sequence $(1/a_n)$ converges to $1/A$
Attempted proof - Using these two theorems above we can write $(a_n b_n/a_n) = (b_n)$ which converges to $B$.
I don't think this is complete can someone help me add details?
 A: Your proof is correct but I'd recommend justifying that $a_n \neq 0$ for all $n$ sufficiently large. Note that this is necessary to make sense of the expression
$$
b_n = \left( \frac{a_nb_n}{a_n} \right). 
$$ 
To justify that $a_n \neq 0$ for all $n$ sufficiently large, we will use that $A = \lim{a_n} \neq 0$. Because $a_n \to A$, there exists $N \in \mathbb{N}$ such that
$$
|a_n - A| < \epsilon = \frac{|A|}{2}
$$
for all $n \geq N$. Therefore, for all such $n$,
\begin{align*}
|A| - |a_n| \leq \left\vert a_n - A \right\vert< \frac{|A|}{2}
\end{align*}
whence $$
0 < \frac{|A|}{2} < |a_n|, \quad \forall n\geq N.
$$
This means that the $N$-tail of $(a_n)$ is never zero and consequently the sequence 
$$
\frac{b_n}{a_n}
$$
is well defined for all $n \geq N$ (and we only really care about the tail of a sequence when discussing limits). Then, by invoking the theorem and lemma you've cited, it follows that
\begin{align*}
\lim{b_n} = \lim\left(\frac{a_n b_n}{a_n}  \right) 
= \frac{\lim(a_nb_n)}{\lim{a_n}} 
= \frac{AB}{A} = B.
\end{align*}
A: Note that if $a_n \to a$ then the sequence is bounded, that is, there is some $L$ such that $|a_n| \le L$.
Suppose $b_n \to b$ then $|a_nb_n-ab| \le |a_n b_n -a_n b| + |a_n b - ab| \le L|b_n-b|+|b||a_n-a|$. Hence $a_nb_n \to ab$.
Suppose $a \neq 0$. Since $a_n \to a$, we see that $|a_n| \ge {1\over 2} |a|$ for $n$ sufficiently large (and hence non zero). Then $|{1 \over a}  - {1\over a_n}| = {|a_n-a| \over |a a_n|} \le {2 \over |a|^2} |a_n-a|$ and hence ${1 \over a_n} \to {1 \over a}$.
Hence if $a_nb_n \to ab$ and $a_n \to a \neq 0$ then
$a_n b_n {1 \over a_n} = b_n \to ab {1 \over a} = b$.
