I'm not sure if I should post it here or in StackOverflow, but anyway...
Prove that: $n^5-2\log{n}=\Omega{(n^5)}$.
Proof:
We need to find $c, n_0 \geq0$ such that, for all $n \geq n_0$, $n^5-2\log{n} \geq cn^5$.
$n^5-2\log{n} \underset{\small{\log{n} < n^2}}{\geq} n^5-2n^2 \geq cn^5$
dividing by $n^5$ (assuming $n>0$):
$c \leq 1-\frac{2}{n^3}$
and since $c \leq \frac{3}{4} \leq 1-\frac{2}{n^3}$ $\forall{n \geq 2}$ (should I prove this here?)
We can choose $n_0=2$ and $c=\frac{1}{4}$, for example.
My question is if this proof is correct, and if there is another (easier) way to do it.
Thanks you.