Is $n^3$ an asymtotically tight bound of $(n^{2.99}).(\log_2n)$? How?

Question

Is $n^3$ an asymtotically tight bound of $(n^{2.99}).(\log_2n)$? If so then how?

Answer 1

No. $n^{2.999}\log_2n$ is tighter, for example.

Answer 2

$g(n)$ is an asymptotically tight bound of $f(n)$ if $f(n) = O(g(n))$, but $f(n) ≠ o(g(n))$. In this case, $n^{2.99} \log n = o (n^3)$. For very large n, $n^{0.01}$ grows faster than $\log n$.