Finding constants C and k for big-O of fraction of polynomials
It works if the denominator doesn't have negative coefficients. Consider, however,
$$ \frac{x}{x-1}. $$
As $x$ tends to infinity, this is $O(1)$. However, your method would give
$$ \frac{x}{x-1} = \frac{1}{1-\tfrac{1}{x}} \leq \frac{1}{1-1} = \infty, $$
which is not very helpful. The problem is that when $x = 1$, the denominator is not positive. In this case it is zero, but for an example like
$$ \frac{x}{x-2} = \frac{1}{1-\tfrac{2}{x}} \leq \frac{1}{-1} = -1, $$
the inequality is in fact wrong.

Often when considering $O(f(x))$, we assume that $f(x)$ is always positive. But sometimes it isn't, and then the assumption that we're making is that $f(x)$ is eventually positive. This is not usually explained when big O notation is defined, but it's the reason behind the extra quantifier "for large enough $x$" which appears in the full definition of big O; this quantifier is only needed when $f(x)$ is not always positive but only eventually positive.

The way to fix the argument is now clear. If $f(x)$ is eventually positive, then take some $y$ such that $f(x) &gt; 0$ for all $y \geq x$, and now apply the method. For example, if $f(x) = x/(x-1)$ then any $y &gt; 1$ would do, say $y = 2$. In that case we have
$$ f(x) = \frac{1}{1-\tfrac{1}{x}} \leq \frac{1}{1-\frac{1}{2}} = 2 $$
for $x \geq 2$.

More generally, suppose that $f(x) = C p(x)/q(x)$, where $p,q$ are monic polynomials (leading coefficient is $1$) and are eventually positive, and $C &gt; 0$. Then for all $\epsilon &gt; 0$ there is $y$ such that for all $x \geq y$,
$$ (C-\epsilon) x^{\deg p - \deg q} &lt; f(x) &lt; (C+\epsilon) x^{\deg p - \deg q}. $$
(Exercise.)