Suppose ${a_{n}}$ and $b_{n}$ are sequence such that $\lim_{\infty }a_{n}=\lim_{\infty }b_{n}=0$ Prove that $\lim a_{n}b_{n}=0$ Please check my proof
$$|a_{n}b_{n}-0|<\epsilon $$
$$ a_{n}b_{n} <\epsilon $$
$$ \frac{1}{\epsilon }< \frac{1}{a_{n}b_{n}} \leftrightarrow  a_{n}b_{n}$$
Choose $a_{N}b_{N}\geq \frac{1}{\epsilon }$ for every $a_{n}b_{n}\geq a_{N}b_{N}$
there fore 
$$\frac{1}{\epsilon }< \frac{1}{a_{n}b_{n}}< \frac{1}{a_{N}b_{N}}$$
the limit equal 0 
 A: $b_n\to 0$ implies that $|b_n|\le 1$ for all $n$ large. Accordingly,
$$
|a_nb_n|=|a_n||b_n| \le |a_n| \to 0.
$$
A: Maybe this answer can help:
Let $\epsilon >0$. Using the assumptions, we can find $N\in\mathbb{N}$ such that if $n\geq N$ then
$|a_n|<\sqrt{\epsilon}$ and $|b_n|<\sqrt{\epsilon}$. Thus, if $n\geq N$ then
$$|a_nb_n-0|=|a_n||b_n|<\sqrt{\epsilon}\cdot\sqrt{\epsilon}=\epsilon.$$
The result follows.
A: Your problem is direct application of following simple problem:
Problem: Suppose $(x_n)$ and $(y_n)$ be two sequences such that $(x_n)$ is bounded and $(y_n) \to 0$. Then $x_ny_n \to 0$ as $n \to \infty$
Solution: Use Squeeze Theorem.
A: Remarks on your proof:
Line 3 does not hold for negative $a_n b_n$. Work wih $\lvert a_n b_n\rvert$ rather.
Line 4 must focus on a choice for $N$.
Line 4 must require $a_n b_n > a_N b_N$ for the second inequality of line 6 to be true.
Suggested proof:
For any challenge $\epsilon>0$ we have $\sqrt{\epsilon}>0$ as well.
As 
$$
\lim_{n\to\infty}a_n = \lim_{n\to\infty} b_n = 0
$$ 
we have $M, N \in \mathbb{N}$ such that for all $n > M$ we have $\lvert a_n\rvert < \sqrt{\epsilon}$ and for all $n > N$ we have $\lvert b_n\rvert < \sqrt{\epsilon}$.
We choose $P = \max \{ M, N \}$ then for all $n > P$ we have
$$
\lvert a_n b_n \rvert
= \lvert a_n\rvert \lvert b_n \rvert 
< \sqrt{\epsilon} \sqrt{\epsilon} = \epsilon
$$
so we have 
$$
\lim_{n\to\infty} a_n b_n = 0
$$
as well.
A: Here is my attempt to retain the structure of the OP's proof and point out omitted details and mistakes.
Two issues with the original proof:


*

*Be careful of jumping to $a_n b_n < \epsilon$ because $a_n b_n$ could be negative. For example, if $a_n b_n = -n$, then this is clearly less than any $\epsilon$ but does not converge.

*There is subtlety in "choosing $a_Nb_N$" because these are two different sequences that might converge to $0$ at different rates. Just because $a_n < \epsilon$ for $n$ greater than some $N$ does not mean $b_n$ has reached the same level of closeness to $\epsilon.$


Fix $\epsilon > 0$. We want to show that there exists an $N$ such that $|a_nb_n - 0| < \epsilon$ for all $n \ge N.$ Notice that
$$
|a_nb_n - 0| = |a_nb_n| = |a_n||b_n|.
$$
Since $a_n \rightarrow 0$, there exists an $N_1$ such that $|a_n| < \sqrt{\epsilon}$ for all $n \ge N_1.$ Similarly, since $b_n \rightarrow 0$, there exists an $N_2$ such that $|b_n| < \sqrt{\epsilon}$ for all $n \ge N_2$. 
Choose $N := \max{N_1,N_2}.$ Then for all $n \ge N$,
$$
|a_n b_n - 0| = |a_n||b_n| < \sqrt{\epsilon}\sqrt{\epsilon} < \epsilon
$$which gives our result. 
