Prove or Disprove Θ

Question

I want to prove or disprove that $3n^3 +n^2\log(n) = Θ(n^3)$. I'm aware that I will need to either prove or disprove both big-o and big-Ω to prove or disprove Θ. I am simply struggling to do so. Help appreciated.

Answer 1

We have that

$$n^3<3n^3 +n^2\log(n)<4n^3$$

Asymptotically, since $\log(n)<n$, the left inequality shows $\Omega(n^3)$ and the right shows $O(n^3)$

Answer 2

Hint: letting $a_n = \frac{3n^3 + n^2\log n}{n^3}$, you have $$a_n \xrightarrow[n\to\infty]{} 3$$ so, as a convergent sequence, the (positive) sequence $\left(a_n\right)_{n\geq 1}$ is bounded. (Even more, for some $n_0$ big enough, you can also say that $2.9 \leq a_n \leq 3.1$ for all $n\geq n_0$.)