Answer
Please see below.
Work Step by Step
We want to prove that $$\lim_{x \to c}f(x)g(x)=0$$ by using $\epsilon - \delta$ definition; that is, we must show that for each $\epsilon >0$, there exists a $\delta >0$ such that $|f(x)g(x)-0|< \epsilon$ whenever $|x-c|< \delta$.
By the assumption, $\lim_{x \to c} f(x) =0$; that is, for any $\epsilon > 0$, there exists a $\delta >0$ such that$$|x-c|< \delta \quad \Rightarrow \quad |f(x)-0|< \epsilon .$$So by choosing $\epsilon = \frac{\epsilon }{M}>0$, $M>0$ is a constant, we conclude that there exists $\delta _1 >0$ such that$$|x-c|< \delta _1 \quad \Rightarrow \quad |f(x)-0|=|f(x)|< \frac{\epsilon }{M}.$$ This means that for any $x \in (c- \delta _1 , c+ \delta _1 )-\{ c \}$ we have $$|f(x)|< \frac{ \epsilon }{M}.$$Also, according to the assumption in the question, in this region, $x \in (c- \delta _1 , c+ \delta _1 )-\{ c \}$, we have $$|g(x)|< M.$$ By combining the last two equations we have the fact that for any $x \in (c- \delta _1 , c+ \delta _1 )-\{ c \}$, we have $$|f(x)||g(x)| < \frac{\epsilon }{M} \epsilon = \epsilon .$$
