Show if $n \sqrt n \in O(n^4)$ I'm quite confused as to how to solve this step by step. What I have done was graph this using Desmos graphing software to visually see it, and I can see that the $n \sqrt n $ is below $n^4$.
I thought I could do something like this:
$n \sqrt n \lt n^4$
$ \sqrt n \lt n^3$
$n^{1/2} \lt n^4$
$1 \lt n^4 / n^{1/2}$
$1 \lt n^{7/2}$
So based on this, $n \sqrt n$ is not bigO of $O(n^4)$. Is that correct? Or would I say it is BigO iff $n > 1$ ?
Thank you.

${{n\sqrt n}\over n^4}={1\over n^{5/2}}$ and $\lim_\limits{n\rightarrow +\infty}{1\over n^{5/2}}=0$.
