If $f$ is bounded and $\lim_{x \to c} g(x) = 0$, then $\lim_{x \to c}g(x)f(x) = 0$ 
Let $g:A\rightarrow \mathbb{R}$ and let $f$ be a bounded function on $A$. Show that if $\displaystyle\lim_{x \rightarrow c}g(x)=0$, then $\displaystyle\lim_{x \rightarrow c}g(x)f(x)=0$ as well.

Since $\displaystyle\lim_{x \rightarrow c}g(x)=0$, we know that $\forall \epsilon>0 $ $\exists \delta >0 $ so that whenever $0<|x-c|<\delta$, it follows that $|g(x)-L| < \epsilon$. And $f(x)$ is bounded, then there is a number $M>0$ satisfying $|f(x)|\leq M$ $\forall x \in A$
$f(x)$ is bounded it means all sequences in $f $ are less than $M$, so it doesn't seem like a stretch to say that 
$|g(x)f(x) - L| < \epsilon$, where, as the problem stated, $L=0$
Then $|g(x)| < \frac{\epsilon} {M}$, since $M$ is an upperbound for $f$
And I'm not quite sure what to do next, any help would be greatly appreciated.
 A: A slightly different approach from the other answer; posted because it is in a similar direction to what you tried:
Since $f$ is bounded on $A$, there exists $M>0$ so that $|f(x)| \leq M$ for $x\in A$ (note that if $f(x)= 0$ for every $x$, the statement is trivial). 
Pick $\varepsilon > 0$. Since $\lim_{x\to c}g(x) = 0$, then there is $\delta > 0$ so that if $0 <|x-c| <\delta$, then $|g(x)| < \varepsilon/M$. Then, since, in particular, $|f(x)|\leq M$ on $(c-\delta,c+\delta)$, we have:
$0 <|x-c|<\delta$ $\Longrightarrow$ $|g(x)f(x)| = |g(x)|\cdot |f(x)| < \left(\varepsilon/M\right)\cdot M = \varepsilon$.
Since $\varepsilon > 0$ was arbitrary, it follows that $\lim_{x\to c} f(x)g(x) = 0$.
A: Note that the function $f$ does not need to be bounded on its entire domain $A$. It is sufficient to assume that it is bounded in some (deleted) neighborhood of $c$ and my answer is based on this idea.

I think OP is interested in a solution based on $\epsilon, \delta$ definition of limit. Before giving a formal proof using Greek symbols it is best to state the argument in language of communication (English here).
Since $f$ is bounded as $x \to c$, it means that the values of $f$ do not exceed some fixed constant $K$. Hence the values of $fg$ do not exceed $K$ times the values of $g$. Since $g$ tends to $0$ as $x \to c$ it follows that the $K$ times $g$ also tends to $0$ and $fg$ does not exceed $Kg$ so it also tends to $0$.
Now we translate the above argument into Greek symbols (which unfortunately is preferred by many instructors). Since $f$ is bounded as $x \to c$ there is a number $K > 0$ and a number $\delta_{1} > 0$ such that $|f(x)| < K$ for all values of $x$ with $0 < |x - c| < \delta_{1}$. We are given that $g(x) \to 0$ as $x \to c$ and hence given any arbitrary number $\epsilon > 0$ there is another number $\delta_{2} > 0$ such that $|g(x)| < \epsilon / K$ whenever $0 < |x - c| < |\delta_{2}$.
Let $\delta = \min(\delta_{1}, \delta_{2})$ and then we have for $0 < |x - c| < \delta$ $$|f(x)g(x)| = |f(x)||g(x)| < K \cdot\frac{\epsilon}{K} = \epsilon$$ and hence $f(x)g(x) \to 0$ as $x \to c$.
A: For this answer, credit must be given to Bungo, the user who edited this question.
$f$ bounded hence $-M \leq f(x) \leq M$ for $M$ large enough. 
We must squeeze $f(x) \cdot g(x)$ between two functions in terms of these bounds. We can do this by noting
$m(x) = \min\{-Mg(x), Mg(x)\} \leq f(x) \cdot g(x) \leq \max\{-Mg(x), Mg(x)\} = n(x)$. Note that these inequalities are satisfied because $|f(x) \cdot g(x)| \leq |Mg(x)|$. 
Note $n(x)$ and $m(x)$ both tend to $0$ as $x \to c$ clearly. Thus, by the squeeze theorem, we may conclude that $f \cdot g \to 0$ as $x \to c$. 
