How to prove that $\lim_{x \rightarrow -\infty}f(x)g(x) = -\infty$ Consider $\lim_{x \rightarrow -\infty}f(x) = \infty, \lim_{x \rightarrow -\infty}g(x) = -\infty$.
How to prove that $$\lim_{x \rightarrow -\infty}f(x)g(x) = -\infty$$ (Without using arithmetic of infinite limits and stuff..just from definition.)?  
So far, by definition, we have:
1.$\forall M_1 >0  ,\exists x_0 ,\forall x<x_0 , f(x)>M_1$.
2.$\forall M_2 >0  ,\exists x_1 ,\forall x<x_1 , g(x)<M_2$
I choose arbitrary $M>0$. and i evaluate this:
$f(x)g(x) < f(x)M_2$ but from here i really stuck. what i need to show? can someone guide me? tnx!
 A: Pick any $A > 1$, set $B=\sqrt{A}$. By definition, there exists $x_0,x_1$ such that what you wrote above holds for $B$ and $-B$ instead of $M_1$ and $M_2$:


*

*$\exists x_0 ,\forall x<x_0 , f(x)>B$.  

*$\exists x_1 ,\forall x<x_1 , g(x)<-B$  


Take $x^\ast\stackrel{\rm def}{=}\min(x_0,x_1)$. Then for $x < x^\ast$, $f(x)g(x) < B\cdot(-B) = -A$. 
A: You have to show that $\forall M\exists x_0\forall x<x_0 f(x)g(x)<M$ (*). 
For this you can choose $M_1=1$ for 1. and have $f(x)g(x)<g(x)$ and use 2. to prove (*).
A: You know that, for all $M>0$, there is $x_0$ such that

if $x<x_0$, then $f(x)>\sqrt{M}$ and $g(x)<-\sqrt{M}$

(why?). Rewrite the second one as $-g(x)>\sqrt{M}$. Then, for $x<x_0$ you have
$$
-f(x)g(x)>M
$$
or
$$
f(x)g(x)<-M
$$
(inequalities involving only positive numbers can be multiplied).

The standard definition for $\lim_{x\to-\infty}f(x)=\infty$ is

for all $M$ there is $x_0$ such that $x<x_0$ implies $f(x)>M$.

In this case it's not restrictive to assume $M$ greater than any constant (for instance $M>0$), because if the inequality $f(x)>M$ holds for $x\in(-\infty,x_0)$, also $f(x)>M_1$ holds in the same interval (but possibly in a larger set) for any $M_1<M$. What we're interested in is that some set of the form $(-\infty,x_0)$ consists of solutions of $f(x)>M$.
The definition for $\lim_{x\to-\infty}g(x)=-\infty$ is very similar:

for all $M$ there is $x_0$ such that $x<x_0$ implies $g(x)<M$.

Here it's not restrictive to assume $M$ less than any constant (for instance $M<0$).
You can easily prove that
$$
\lim_{x\to-\infty}g(x)=-\infty
\quad\text{if and only if}\quad
\lim_{x\to-\infty}(-g(x))=\infty
$$
