$f(x) \geq g(x) \Leftarrow \lim_{x \rightarrow \infty}\frac{g(x)}{f(x)}=0 $?

Question

I want to know if function $f(x)$ is greater or equal than $g(x)$. If I prove that

$\lim_{x \rightarrow \infty}\frac{g(x)}{f(x)}=0$ then is it so?

Answer 1

If $$ \lim_{x\to\infty}\frac{g(x)}{f(x)}=0, $$ then we have that for any $\varepsilon>0$ $$ \biggl|\frac{g(x)}{f(x)}\biggr|<\varepsilon $$ when $x>x_0$ for some $x_0\in\mathbb R$. Set $\varepsilon=1$. Then we obtain $$ \biggl|\frac{g(x)}{f(x)}\biggr|<1 $$ or $$ |g(x)|<|f(x)| $$ when $x>x_0$ for some $x_0\in\mathbb R$. Hence, function $|f(x)|$ is greater than $|g(x)|$ when $x$ is sufficiently large. However, it does not hold in the opposite direction as noted in the comments.

Answer 2

This is certainly not an "$\Leftrightarrow$", for example $f(x)=g(x)=1$ leds to $\lim_{x\to\infty}\frac{g(x)}{f(x)}=1$.

We do have the following related results:

• If $f(x)\ge g(x)>0$ for all $x$ (or just for all $x>x_0$) then $\limsup _{x\to\infty}\frac{g(x)}{f(x)}\le1$

• If $\left|\lim_{x\to\infty}\frac{g(x)}{f(x)}\right|<1$ then there exists $x_0$ such that $|f(x)|\ge |g(x)|$ for all $x>x_0$