Question about growth rates of functions involving n and logn
There exists some $n_0 &gt; 0$ such that $\log n \le n^{0.0001}$ for all $n \ge n_0$.
This can be seen, e.g., by taking the limit:
$$
\lim_{n \to \infty} \frac{\log n}{n^{0.0001}} = 
\lim_{n \to \infty} \frac{10000}{ n^{0.0001}} = 0.
$$

This means that, for $n \ge n_0$,
$$
f(n) = \log^{100} n + n^{0.01} \le  \left( n^{0.0001} \right)^{100} + n^{0.01} = 2n^{0.01}.
$$

Therefore, using the fact that $g(n) \ge n^{0.05}$:
$$
\lim_{n \to \infty} \frac{f(n)}{g(n)} \le  
\lim_{n \to \infty} \frac{2n^{0.01}}{n^{0.05}} = 
\lim_{n \to \infty} \frac{2}{n^{0.04}} =
0,
$$
implying that $f(n)=o(g(n)).$