If $f(x) \rightarrow -\infty$ and $g(x) \rightarrow -\infty$ does $f(x)\cdot g(x) \rightarrow \infty$?

Question

As the question states:

When the process is same for all statements, can we assume that

if $f(x) \rightarrow -\infty$ and $g(x) \rightarrow -\infty$ does $f(x)\cdot g(x) \rightarrow \infty$?

I can't think of any possible functions, where this does not hold, but I want to be sure (infinities can be tricky).

Suppose $a\in\mathbb{R}$. What is the meaning of $f(x)\to -\infty$ as $x\to a$? — Arian, Sep 29 '17 at 16:11
This one is one of the non-tricky aspects of infinity. The limit is indeed $\infty$. — Paramanand Singh, Sep 29 '17 at 16:12

score 1 · Accepted Answer · answered Sep 29 '17 at 16:11

This argument assumes you are using "as $x \to \infty$"". It can "easily" be modified to conform to other cases.

If $f(x) \to -\infty$ then for every $u > 0$ there exists an index $N_1$ such that $x > N_1$ implies $f(x) < -u$.

If $g(x) \to -\infty$ then for every $v > 0$ there exists an index $N_2$ such that $x > N_2$ implies $g(x) < -v$.

Let $w$ be any positive real number. Then we can write $w = (-t) \cdot (-t)$ where $t = \sqrt w > 0$.

Hence there exists positive indexes $N_1$ and $N_2$ such that $x > N_1$ implies $f(x) < -t$ and $x > N_2$ implies $g(x) < -t$.

Let $N = \max\{N_1, N_2\}$. Then

\begin{align} x > N &\implies \text{$f(x)<-t \ $ and $ \ g(x)<-t$} \\ &\implies f(x)g(x) > t^2 \\ &\implies f(x)g(x) > w \end{align}

It follows that $f(x)g(x) \to \infty$

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