big o - Big O complexity with n^2 log(n) -

two questions:

first, if f(n) = n(3n + nlog(n)) why f(n) Ω(n²)?

second, why n²log(n) not o(n²)?

these both consequences of fact log(n) tends infinity n tends infinity.

1) n(3n + nlog(n)) omega(n^2) because large n 3n negligible , n^2log(n) bounded below n^2

2) n^2log(n) not o(n^2) since, constant k > 0, n > e^k have n^2log(n) > kn^2, no k satisfies n^2log(n) < kn^2 finitely many n.