limit of product exists and one limit exists Question is to check : 
If  $\lim_{n\rightarrow \infty}a_nb_n$ exists and $\lim_{n\rightarrow \infty}a_n$ exists implies $\lim_{n\rightarrow \infty}b_n$ exists.
Considering $a_n=\frac{1}{n}$ and $b_n=n$ then we see that  $\lim_{n\rightarrow \infty}a_nb_n$ exists, equals to $1$  and
$\lim_{n\rightarrow \infty}a_n$ exists and equals to $0$. In this case $\lim_{n\rightarrow \infty}b_n$  does not exists..
So, the answer to the question is Not always..
Now, what if $\lim_{n\rightarrow \infty}a_n$ exists and is non zero and $(b_n)$ is bounded?
Suppose that $\lim_{n\rightarrow \infty}a_nb_n=M$ with  $\lim_{n\rightarrow \infty}a_n=P\neq 0$ and $|b_n|\leq A$ for all $n\in \mathbb{N}$.
I claim that $\lim_{n\rightarrow \infty}b_n=\frac{M}{P}$
Consider $|b_n-\frac{M}{P}|$.. We estimate this. Given $\epsilon>0$ there exists $N\in \mathbb{N}$ such that $|a_nb_n-M|<\epsilon$ and $|a_n-P|<\epsilon$ for all $n\geq N$. 
$$|b_n-\frac{M}{P}|=\frac{1}{P}|Pb_n-M|=\frac{1}{P}|Pb_n-a_nb_n+a_nb_n-M|\leq \frac{1}{P}|b_n||a_n-P|+\frac{1}{P}                                   \epsilon$$
As $(b_n)$ is bounded, we have for all $n\geq N$
$$|b_n-\frac{M}{P}|\leq \frac{1}{P}A\epsilon+\frac{1}{P}                                   \epsilon=\epsilon\left(\frac{1}{P}(A+1)\right)$$
Thus, we are done.
I am just wondering if i can relax any of the conditions that i have assumed. Help me to know more about this.

If 
$$\;\lim_{n\to\infty}a_n=L\neq0\;,\;\;\lim_{n\to\infty}a_nb_n= K\;,\;\;\text{then since for almost all indexes}\;\;a_n\neq0\,,$$
we get that for all indexes except a finite number of them, from arithmetic of limits:
$$b_n=\frac{a_nb_n}{a_n}\xrightarrow[n\to\infty]{}\frac KL$$
and all this is well-defined and always finite since $\;L\neq0\;$ . No need to require a priori boundedness for $\;\{b_n\}\;$ .
