STEPS: Proving $\displaystyle \lim_{x\to a} f(x)g(x) = -\infty$ Given $\displaystyle \lim_{x \to a} f(x) = \infty$ and $\displaystyle\lim_{x \to a} g(x) = c$ where $c<0$.
Prove that $\displaystyle \lim_{x \to a} f(x)g(x) = -\infty$ only using the precise definitions of limit and infinite limit.
I get the intuitive idea. But... how do we actually write it out?
 A: These exercises are almost always solved in a standard way.


*

*Write down the assumptions, using their definition. For instance, $\lim_{x \to a} f(x)=\infty$ means: for every $M>0$ there exists $\delta >0$ with the property that $0<|x-a|<\delta$ implies $f(x)>M$. Now please write down the meaning of your second assumption about $g$.

*Write down what you need to prove. In you case, you need to prove that, for every $M>0$ there exists $\delta >0$ such that $0<|x-a|<\delta$ implies $f(x)g(x)<-M$.

*Try to use the information provided by the assumptions to conclude.


To give you even more hints, you'll almost surely have to play with two numbers $\delta_1$ and $\delta_2$ coming from the assumptions. You'll probably want to select $\delta = \min \{\delta_1,\delta_2\}$.
A: Let $A>0$ and $\epsilon=-\frac c2>0$ then there's $\delta_1,\delta_2>0$ such that 
$$|x-a|<\delta_1\implies f(x)>A$$
and
$$|x-a|<\delta_2\implies \frac{3c}2<g(x)<\frac c2$$
Now we have
$$|x-a|<\delta=\min(\delta_1,\delta_2)\implies f(x)g(x)<\frac{Ac}2$$
and this for arbitrary $A$. We conclude.
