I was asked the following:
We are given the functions $f(n)=n^{10\log(n)}$ and $g(n)=(\log (n))^n$.
Which of the following statements is true:
$f(n)\in\mathcal{O}(g(n))$,
$f(n) \in \mathcal{\Theta}(g(n))$,
$f(n) \in\mathcal{\Omega} (g(n))$
According to wolfram, the first result is true. This is because:
$$\lim_{n \to \infty} \frac{f(n)}{g(n)} =0$$ which implies $f(n)\in\mathcal{O}(g(n))$.
Here is what I did:
$f(n) = n^{10 \log(n)} = (2^{ \log (n)})^{10 \log (n)}=2^{10 (\log(n))^2}$
For $n \geq 5$, $\log (n) >2$, and so:
$2^{10 (\log(n))^2} < (\log (n))^{10 (\log(n))^2}$
notice that $$\lim_{n \to \infty} \frac{10 (\log(n))^2}{n}=0$$ (apply lhopital twice) and so if $n$ is large enough, $10 (\log(n))^2 < n$ and we can get
$$(\log (n))^{10 (\log(n))^2} < (\log(n))^{n}$$
proving the thing we wanted to prove.
My problem is that I used calculus. If possible, I would rather not use limits. I actually want to find such $n_0$ and $c$ such that $f(n) < cg(n)$ for all $n >n_0$. I did not do that in this "proof". Would appreciate help.