Continuous Functions and Their Product 
Consider two functions $f,g:R\rightarrow R$. Suppose that $f$ is continuous at $0$ and $f(0)=0$; $g(x)$ is bounded but may not be continuous at $0$.
Prove that $fg$ is continuous at $0$.

So far I have the definition of what it means for $f$ to be continuous at $0$. Where is the best place to go from here?
 A: First figure out what is it exactly that you need to prove. Continuity of $f\cdot g$ at $0$ means that the limit $\lim_{x\to 0}f(x)\cdot g(x)$ exists and equals $f(0)\cdot g(0)$, which, since $f(0)=0$, is simply $0$. So, you need to show that $\lim_{x\to 0}f(x)\cdot g(x)=0$. Now you can use the $\epsilon$-$\delta$ definition to establish that. You know that $g$ is bounded by some constant $M>0$ and that $\lim_{x\to 0}f(x)=0$. So, given any $\epsilon>0$, consider using $\frac{\epsilon}{M}$ in the definition of limit for $\lim_{x\to 0}f(x)=0$. Done correctly, this should give you the desired proof. 
A: Here $g(x)$ is bounded. so there  is an $ \alpha  \in \mathbb{R}\ s.t. \ g(x)$<$\alpha \implies |g(x)-g(0)|<2\alpha $ , 
since $f$  is continuous at $x=0$, then $|f(x)-f(0)|<\frac{\epsilon}{\alpha} \forall |x|< \delta\ $
$|f(x)g(x)-f(0)g(0)|=|f(x)g(x)-f(0)g(x)+f(0)g(x)-f(0)g(0)| \\ \ \leq |g(x)||f(x)-f(0)|+|f(0)||g(x)-g(0)| \\ < \alpha.\frac{\epsilon}{\alpha}+0.2\alpha= \epsilon \ \forall\ |x|< \delta\ $ 
