Is $n^2 \log n$ in $O (n^2)$ am confused I graphed the functions and when $n_0$ is greater than $3$, $cg(n)$ is always greater than $n^2 \log n$ so it would seem to me by definition that $m^2 \log n$ is in $O (n^2) $. I tried to prove it by dividing both sides of inequality by $n^2$ . The $c \ge \log n $. As n tends to infinity there will be values of c that are &gt; than logn . I have to say if $n^2 \log n $ is in $O(n^2)$ . My classmate says it isn't to me it looks like it is.  It is 6.00 am here and my brain is mush so apologies if this is a really trivial questions and you are all laughing at my stupidity. It is a little confusing I admit. It seem to me that $n^2 \log n$ above $n_0= 3$ is always greater than $n^2$
$n^2\log(n)$ isn't $O(n^2)$. You need to have a constant $c$, so since $\log(n)$ tends to infinity, for any fixed $c$ you will choose, at some point $\log(n)$ will pass $c$, that is: $\forall n&gt;n_0: c &lt; \log(n)$ for some $n_0$. This directly means that $\forall n&gt;n_0: cn^2  &lt; n^2 \log(n)$ which means that $n^2 \log(n)$ is asymptotically bigger than $cn^2$.