Answer
$\lim\limits_{n\to\infty} (a_nb_n)=0$
Work Step by Step
Recall the property of the absolute value: $-|a_n| \leq a_n \leq |a_n|$ for all the values of $n$
consider $\lim\limits_{n\to\infty} |a_n|=0$
By the limit laws of a sequences, we have $\lim\limits_{n\to\infty} |a_n|=0$; $\lim\limits_{n\to\infty} -|a_n|=-\lim\limits_{n\to\infty} |a_n|=0$
By the squeeze theorem for the sequence, we have $\lim\limits_{n\to\infty} a_n=0$. Thus, If $\lim\limits_{n\to\infty} |a_n|=0$, then $\lim\limits_{n\to\infty} a_n=0$
Consider a small number $\epsilon \gt 0$ , then $a_n \to 0$
There will exist a number $N$ such that $|a_n| \lt \dfrac{\epsilon }{M}$ for the all values of $n \geq N$
$|a_nb_n|=|a_n||b_n| \lt \dfrac{\epsilon }{M} (M)=\epsilon$
Hence, it has been proved that $\lim\limits_{n\to\infty} (a_nb_n)=0$