Answer
$\lim\limits_{n\to\infty} (a_nb_n)=0$
Work Step by Step
We need to prove that $\lim\limits_{n\to\infty} (a_nb_n)=0$
Apply the property of the absolute value as follows:
$-|a_n| \leq a_n \leq |a_n|$ for all the values of $n$
Let us consider that $\lim\limits_{n\to\infty} |a_n|=0$
Suppose $\lim\limits_{n\to\infty} a_n=L$ and the function $f(a_n)$ is continuous at the limit $L$.
Apply the limit laws of sequences.
$\lim\limits_{n\to\infty} |a_n|=0$; $\lim\limits_{n\to\infty} -|a_n|=-\lim\limits_{n\to\infty} |a_n|=0$
The squeeze theorem for a sequence states that:
$\lim\limits_{n\to\infty} a_n=0$.
This means that when $\lim\limits_{n\to\infty} |a_n|=0$, then we have $\lim\limits_{n\to\infty} a_n=0$
Now, let us consider a small number $\epsilon \gt 0$ , then we have $a_n \to 0$, so that there will exist a number $N$ such that $|a_n| \lt \dfrac{\epsilon }{M}$ when $n \geq N$
This gives us: $|a_nb_n|=|a_n||b_n| \lt \dfrac{\epsilon }{M} (M)=\epsilon$
Hence, it has been verified that $\lim\limits_{n\to\infty} (a_nb_n)=0$
