How is $n^{1.001} + n\log n = \Theta (n^{1.001})$? I am studying for an exam and stumbled across this here:
https://cs.stackexchange.com/questions/13451/few-big-o-example
(I cant comment there since commenting needs 50 reps and I am a new user. Thought math exchange would help)
The chosen answer says for large $n$, $n^{0.001}$ is larger than $\log n$. But that does not make sense. $0.001$ is close to $0$. So anything raised to the power $0.001$ should come out to be slightly more than $1$ right?
e.g if n = 1000,000 then $n^{0.001}$ is equal to $1.014$ whereas $\log n$ will be equal to almost 20 if $\log$ has a base of 2.
Where am I thinking wrong?
Is there any other way of showing this relationship?

For intuition, consider something like $n = 2^{10 ^ 6}$. We have $log(n) = 10^6$ and $n^{0.001} = 2^{10^3} \approx 10^{301}$
In general
$\displaystyle \lim_{n\to \infty} \frac{(\log n)^a}{n^b} = 0$ if $b > 0$
