i have $f(n)=n^3+n\log^2n$ and i was trieng to prove that $f(n)=n^3+n\log^2n$ = $\theta(n^3)$. but i feel that i am doing it all wrong , which means i understand why the statent is true but i don't know if the way i proved it was enough.. here is what i did so far :
first i need to prove that : $f(n)=n^3+n\log^2n$ = $O(n^3)$ We can choose constant C like this : if we devide the equation with $n^3$ then we get : $1+n\log^2n/n^3$ $<=$ $C $ but $n\log^2n/n^3$ <= 1 for every n so if we choose C to be C>=2 then this will be proved.
second i need to prove :$f(n)=n^3+n\log^2n$ = $pi(n^3)$ this i think doesn't need to be proved because if i choose any C<1 then for every n we will get : $n^3+n\log^2n$ >= $C(n^3)$
here is
Hint: $\lim_{n \to \infty} \frac{\log n}{\sqrt n} = (\infty / \infty) = (by L'Hospital) \lim_{n \to \infty}\frac{2\sqrt n}{n} =0$