# Proving Big Omega of a polynomial without limits

Here is the definition of $$\Omega$$:

$$f(n) = Ω(g(n))$$ iff there exist positive constants $$c$$ and $$n_0$$ such that $$f(n) \ge cg(n)$$ for all $$n\ge n_0$$.

Here is one theorem:

If $$f(n) = a_m n^m + \cdots + a_1 n + a_0$$ and $$a_m > 0$$, then $$f(n) = \Omega(n^m)$$.

I want to prove this, without using limits. Despite many hours of searching across the internet, all I could find is proofs using limits. Is there any other way?

Define $$M = \max(|a_0|/a_m, |a_1|/a_m, \ldots, |a_{m-1}|/a_m)$$, and take $$c = a_m/2$$ and $$n_0 = 2mM$$. Then for $$n \geq n_0$$, \begin{align*} f(n) &= a_m n^m \left(1 + \frac{a_{m-1}}{a_m} \cdot \frac{1}{n} + \cdots + \frac{a_1}{a_m} \cdot \frac{1}{n^{m-1}} + \frac{a_0}{a_m} \cdot \frac{1}{n^m}\right) \\ &\geq a_m n^m \left(1 - \frac{M}{n_0} - \cdots - \frac{M}{n_0^{m-1}} - \frac{M}{n_0^M}\right) \\ &\geq a_m n^m \left(1 - \frac{mM}{n_0}\right) \\ &= c n^m. \end{align*}