Show that limit of quotient with infinity involved is equal to 0 Given $\lim_{x\to \infty}f(x) = -\infty$ and $\lim_{x\to \infty}g(x) = M$, where $M \neq 0$, I would like to prove that $\lim_{x\to \infty}|g(x)/f(x)| = 0$. 
My attempt was to take $C$ such that for every $x \ge C$ we have $|g(x) - M| < L - M$ (so $g(x) < L$) and $f(x) \lt L/\epsilon$. Unfortunately, I don't think I can say anything about the limit of quotient with these inequalities. How do I proceed?
 A: Let $\varepsilon>0$. By assumption, you can find $C_1>0$ such that, for $x>C_1$,
$$
|g(x)-M|<\varepsilon
$$
The second inequality can be rewritten as
$$
M-\varepsilon<g(x)<M+\varepsilon
$$
which implies
$$
-|M|-\varepsilon<g(x)<|M|+\varepsilon
$$
so $|g(x)|<|M|+\varepsilon$.
Also by assumption there exists $C>0$ (and without loss of generality we can assume $C>C_1$), such that, for $x>C$,
$$
f(x)<-\frac{|M|+\varepsilon}{\varepsilon}
$$
which implies $|f(x)|>(|M|+\varepsilon)/\varepsilon$. Therefore, for $x>C$,
\begin{align}
|g(x)|&<|M|+\varepsilon \\[6px]
\frac{1}{|f(x)|}&<\frac{\varepsilon}{|M|+\varepsilon}
\end{align}
By multiplying the two inequalities, we get that, for $x>C$,
$$
\left|\frac{g(x)}{f(x)}\right|<(|M|+\varepsilon)\frac{\varepsilon}{|M|+\varepsilon}
=\varepsilon
$$

You can see that it is actually immaterial that $g$ has limit for $x\to\infty$: what matters is that it is bounded over some interval $(C_1,\infty)$. This is not different from proving that the product of a bounded function by a function having limit $0$ has limit $0$.
