Let $m, n \in \mathbb{Z}^{\geq 0}$.
Let $a_1, ..., a_n \in \mathbb{R}$.
Let $b_1, ..., b_m \in \mathbb{R}$.
Assume $\displaystyle\frac{a_n}{b_m} > 0.$
Let $f(x) = \displaystyle\frac{a_n x^n + a_{n - 1} x^{n - 1} + ... + a_1 x + a_0}{b_m x^m + b_{m - 1} x^{m - 1} + ... + b_1 x + b_0}$
So far, I have managed to prove that as $x \rightarrow \infty, f(x) \rightarrow \displaystyle\frac{a_n}{b_m} x^{n - m}$.
In other words,
$\forall \varepsilon > 0, \exists x_0 \in \mathbb{R} \textrm{ s.t. } \forall x \in \mathbb{R}, x > x_0 \Rightarrow \bigg| f(x) - \displaystyle\frac{a_n}{b_m} x^{n - m} \bigg| < \varepsilon$.
When defining limits as $x \rightarrow \infty$, only $x$ values greater than zero need to be considered.
$\forall \varepsilon > 0, \exists x_0 > 0 \textrm{ s.t. } \forall x \in \mathbb{R}, x > x_0 \Rightarrow \bigg| f(x) - \displaystyle\frac{a_n}{b_m} x^{n - m} \bigg| < \varepsilon$
Since $\frac{a_n}{b_m} > 0$ and in the case that $x > x_0 > 0$,
$\frac{a_n}{b_m} x^{n - m} > 0$
Therefore, $f(x)$ is eventually always positive.
I would like to prove formally that $\exists M \in \mathbb{R} \textrm{ s.t. } \forall x \in \mathbb{R}, x > M \Rightarrow f(x) > 0.$
What I have tried so far:
I take $\varepsilon = \displaystyle\frac{a_n}{b_m} x^{n - m}$ so I can get a result of $f(x) > 0$.
However, $x_0$ depends on $\varepsilon$ and $M$ depends on $x_0$. This leads to $M$ depending on $x$ which is illegal since $x$ must be declared after $M$.
I need help doing this in the right order.