I'm trying to prove that $n^2 \not\in \omega(2^n)$ and I have to do it from definition.
$f(n) \in \omega(g(n)) = \left\{f(n)| \forall c>0, c \in \mathbb{R}, \exists n_0 \in \mathbb{N}, \forall n \ge n_o (f(n) \ge c g(n)) \right\}$
So we're trying to prove that $n^2 \ge c 2^n$.
Using limits I can prove that $\exists c, n^2 \not\ge c 2^n$ $$\frac{n^2}{2^n} \ge c$$
$$\lim_{x \to \infty} \frac{n^2}{2^n} = 0$$
$$0 \ge c$$
However I am wondering is there a way to prove it without the us of limits?
The result that you're showing is actually stronger: You've proved that $n^2 \in o(2^n)$ which is equivalent to $\lim_{n\rightarrow\infty} n^2/2^n = 0$.
Since these asymptotic notions are defined in terms of limits, you'll most likely will be using limits (at least implicitly) for proving such statements.