If an algorithm has a running time $ T(n) = O(n$ log $n)$, would it be possible to show that $T(n) = o(n^2)$?
2026-04-01 01:58:56.1775008736
Comparing algorithmic complexities
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Yes:
$$ T(n)=O(n\log(n))\implies\lim_{n\to\infty}\frac{T(n)}{n\log(n)}=c\in\mathbb{R}\\ \lim_{n\to\infty}\frac{T(n)}{n^2}=\lim_{n\to\infty}\frac{T(n)}{n\log(n)}\frac{\log(n)}{n}=c*0=0\implies T(n)=o(n^2) $$