Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.4
Solution: Let $\epsilon>0$. Since $\lim s_n=0$, there exists $N>0$ such that
$$
|s_n-0|=s_n < \frac{\epsilon}{M+1}
$$ for all $n>N$. Therefore $n>N$ implies that
$$
|s_nt_n|=|s_n|M<\frac{\epsilon}{M+1}M<\epsilon
$$ for all $n>N$ as desired. Thus $\lim (s_nt_n)=0$.
Here we use $\dfrac{\epsilon}{M+1}$ instead of $\dfrac{\epsilon}{M}$ because $M$ may be zero. With this modification, we do not need to consider the case where $M=0$ separately.