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.
 A: 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 $\;\{b_n\}\;$ .
A: You get the condition that $(b_n)$ is bounded for free. Let us first show that by contradiction. Assume that $(b_n)$ is unbounded. Then there exists subsequence $(b_{p(n)})$ of $(b_n)$ such that $|b_{p(n)}|>n$, for all $n$. But, then we have $$|na_{p(n)} |\leq |a_{p(n)}b_{p(n)}| \leq M$$ where $M$ is such that $|a_nb_n|\leq M$, which exists by convergence of $(a_nb_n)$. It follows that $$0\leq |a_{p(n)}| \leq \frac M n \implies \lim_na_n = \lim_n a_{p(n)} = 0$$ Contradiction.
On the other hand, you could easily prove that $\lim_na_n\neq 0$ implies convergence of $(b_n)$ just by noting the general rule: $$\lim_nb_n\neq 0\implies\lim_n\left(\frac{a_n}{b_n}\right) = \frac{\lim_na_n}{\lim_nb_n}$$ for convergent sequences $(a_n)$ and $(b_n)$.
A: *

*If $\lim_{n}a_{n}=0$, the conclusion is not correct. Take as a counter example:
$$a_{n}=1/n$$
$$b_{n}=\sin(n\pi).$$ By squeeze theorem 
$a_{n}b_{n}\rightarrow0$, but $b_{n}$ has no limit.

*If $\lim_{n}a_{n}\not=0$
Then the result is true. Because, for $n>N$ (ultimately), we have $a_{n}\not=0$ and so we can write 
$$b_{n}=\frac{1}{a_{n}}.a_{n}b_{n}$$ and since the limit of the right side exists, the left side must have also a limit.
