I'm working through a big-O problem and have the intuition to know the answer, but don't feel comfortable in my proof. I need to prove from definitions (i.e. proving that there exists two constants $c$ and $n_0$) that
$$3^{3\log(n) + \log \log(n)}\log(n^{450}) \in O(9^{2\log(n)}).$$
Now I end up selecting an $n_0$ value of 1, and a $c$ value of 1. The entire left side of my equation collapses into $0$. I'm left with $0 \leq 9^0 = 1$, which is true.
Now my questions: this seems far too easy. Am I really allowed to pick those constants myself? Also, the example I used just doesn't seem very rigorous, but it does make it clear how one side grows much faster than the other, does it not? Any advice would be greatly appreciated.
You have checked the inequality only for $n_0$ you need to check it also for all $n>n_0$.
An strategy, among many, could be to divide both functions.
$$\begin{align}\frac{3^{3\log(n)+\log\log(n)}\log(n^{450})}{9^{ 2log(n)}}&=\frac{3^{3\log(n)+\log\log(n)}\log(n^{450})}{3^{4 log(n)}}\\&=450\cdot3^{-\log(n)+\log\log(n)}\log(n)\\&=450\cdot3^{-\log(n)+\log\log(n)}3^{\frac{\log\log(n)}{\log(3)}}\\&=450\cdot3^{-\log(n)+\log\log(n)+\frac{\log\log(n)}{\log(3)}}\end{align}$$
And try to prove this is bounded for large $n$. So far this is the definition. What we can try to do now is to compute the limit as $n\to\infty$. If this limit is exists, as a finite limit, then it follows that the quotient is bounded.
In our particular case we can compute $$\begin{align}\log\log(n)-\log(n)+\frac{\log\log(n)}{\log(3)}&=\log\left(\frac{\log(n)}{n}\right)+\log(\log(n)^{1/\log(3)})\\&=\log\left(\frac{(\log(n))^{1+1/\log(3)}}{n}\right)\end{align}$$
Apply L'Hospital to the fraction inside the first $\log$. We get
$$\left(1+\frac{1}{\log(3)}\right)\frac{(\log(n))^{1/\log(3)}}{n}$$
and again
$$\left(1+\frac{1}{\log(3)}\right)\frac{1}{\log(3)}\frac{1}{(\log(n))^{1-1/\log(3)}n}\to0$$
Therefore the original fraction tends to $450\cdot3^0=450$ and is therefore bounded. From this limit we know that any constant greater than $450$ should work for some large $n_0$.