Is n$^{99}+n^{98}$ $\in$ O($n^{99}$)?

Question

If so, are there witnesses that prove this true? My initial thought is no, considering that the limit of $n^{99}+{n^{98} \over n^{99}}$ as n approaches $\infty$ is 1, however my professor had mentioned this is not sufficient to prove big O relation.

Answer 1

Yes, consider that $n^{99} + n^{98} < 2 \cdot n^{99}$ for $n \ge 2$. Also I don't think there is anything wrong with your method of proof (except that you're getting the wrong conclusion somehow).