Does $\lim\limits_{y\rightarrow -\infty}\frac{g(y)}{f(y)}=0$ imply that $\lim\limits_{y\rightarrow -\infty}\frac{f(y)}{g(y)}$ cannot be finite

Question

Suppose we have two functions $g:\mathbb{R}\rightarrow [0,1]$ and $f:\mathbb{R}\rightarrow [0,1]$ such that
$$\hspace{1cm}\lim\limits_{y\rightarrow -\infty}g(y)=\lim\limits_{y\rightarrow -\infty}f(y)=0 \tag{1}$$
Moreover, we know that
$$\hspace{1cm}\lim\limits_{y\rightarrow -\infty}\frac{g(y)}{f(y)}=0 \tag{2}$$
which I interpret as saying "g goes to zero faster than $f$".

Do (1), (2) imply that
$$
\lim\limits_{y\rightarrow -\infty}\frac{f(y)}{g(y)}
$$
cannot be a finite number? If this is not true, can you make a counterexample?

Answer 1

Yes, it is not finite; if it were finite, say $\lim_{y\rightarrow -\infty}\frac{f(y)}{g(y)}=L$, then this would imply that $$0\cdot L=\left(\lim_{y\rightarrow -\infty}\frac{g(y)}{f(y)}\right)\left(\lim_{y\rightarrow -\infty}\frac{f(y)}{g(y)}\right)=\lim_{y\rightarrow -\infty}\left(\frac{g(y)}{f(y)}\frac{f(y)}{g(y)}\right)=\lim_{y\rightarrow -\infty}1=1.$$