A task from an old exam about Big O notation:
Find a non-negative function f such that $f \in \Omega(n^2)$ and $f \in o(n^2 \sqrt{n})$
I'm not sure if I did it correctly, so I've checked both definitions on the internet and found that
$$\lim_{n \rightarrow \infty} \frac{f(n)}{g(n)} > 0 \Rightarrow f \in \Omega(g) $$
$$\lim_{n \rightarrow \infty} \frac{f(n)}{g(n)} = 0 \Rightarrow f \in o(g)$$
So we need to find a function $f$ which satisfies the following condition:
$$\lim_{n\rightarrow \infty} \frac{f(n)}{n^2} > 0 \text{ and } \lim_{n \rightarrow \infty} \frac{f(n)}{n^2\sqrt{n}} = 0 $$
If we choose $f(n) = 2n^2$, the first condition will be satisfied because $\frac{2n^2}{n^2} = 2 > 0$
The second condition should be satisfied too since $\frac{2n^2}{n^2\sqrt{n}} = \frac{2}{\sqrt{n}}$ and let $n$ go towards $\infty$, we end up with zero.
I hope it is correct like that?