Big Omega problem : is $n^2\in\Omega (2n^2)$?

Question

Is $n^2\in\Omega (2n^2)$? If we find the limit we can see $\frac{1}{2}>0$, which means it is true, but I haven't learned the limit method. I need to figure out using this definition $\exists c>0,k≥0:\forall n>k:f(n)\geq cg(n)$ that $n^2\in\Omega(2n^2)$...

Answer 1

Let $c = \frac{1}{2}$ and $k = 0$. Then $f(n) = n^{2} \geq \frac{1}{2} \cdot 2n^{2}$ for all $n \geq 0$. Since $\frac{1}{2} \cdot 2n^{2} = f(n)$, this proves the result.