Given a function $f$ such that $\lim \limits_{x \to \infty} f(x) =0$, and want to see if $f(x) >0 $ or $f(x)<0$, but its so hard to tell(its very complicated function).
My approach to solve this problem is : find another function $g$ such that $g(x)$ is very big function and positive and multiply it with $f(x)$ and if the $\lim \limits_{x \to \infty} f(x)g(x) = \infty$ then $f(x)>0$ for all sufficiently large number and if the $\lim \limits_{x \to \infty} f(x)g(x) = -\infty$ then $f(x)<0$ for all sufficiently large number.
Is this approach valid ?! which means it does tell you if $f(x)>0$ or $f(x)<0$ for all sufficiently large numbers ?
We assert validity with proofs.
Notice
By definition of limits, we have that $\lim\limits_{x \to \infty} f(x) g(x) = \infty $ is equivalent to saying:
$\forall M >0 \exists X$ such that $x \ge X \implies f(x)g(x) > M \iff f(x) > \frac{M}{g(x)}$
thus for $f(x) >0$ it's sufficient to have $g(x) > 0$. If your choice of $g(x) >0$ then your derivation is valid.
Notice $f(x) > \frac{M}{g(x)}$ means that $f(x)$ can approach 0, for $g(x) \to \infty$, but will do it from the positive side.
Now the negative case.
By definition of limits, we have that $\lim\limits_{x \to \infty} f(x) g(x) = -\infty $ is equivalent to saying:
$\forall M >0 \exists X$ such that $x \ge X \implies f(x)g(x) < -M \iff f(x) < -\frac{M}{g(x)}$
thus for $f(x) < 0$ it's sufficient to have $g(x) > 0$ just like in the positive case.
Notice $f(x) < -\frac{M}{g(x)}$ means that $f(x)$ can approach 0, for $g(x) \to \infty$, but will do it from the negative side.