What is the growth rate and how can we order given functions w.r.t increase in growth rate? How can we order the following functions ascending with respective to the rate of growth:
$n^n$,
$nlog_2n+n\sqrt{n}$,
$3n^3+5n^2+6$,
$n!+2^n$,
$n+log_2(log_2n)$,
$n^{20}+2^n$,
$log_2n$,
$2^n$,
5,
$log_2(log_2n)$.
We say that a function $f(x)$ has a greater rate of growth than $g(x)$ if
$\lim \limits_{x\to\infty} \frac{f(x)}{g(x)} \to \infty$
So, one way to order some functions with respect to the rate of growth is to compute that limit. For example, if we want to see which of the first two functions growths faster, you should compute:
$\lim \limits_{n\to\infty} \frac{n^n}{nlog{n}+n\sqrt{n}} \hspace{2pt} \underset{\mathrm{Hôpital}}{=} \hspace{2pt} \lim \limits_{n\to\infty} \frac{n^n(log{n}+1)}{\frac{3}{2}\sqrt{n}+log{n}} \hspace{2pt} \underset{\mathrm{Hôpital}}{=} \hspace{2pt} \lim \limits_{n\to\infty} \frac{n^{n-1}+n^n(log{n}+1)^2}{\frac{1}{4\sqrt{n}}+\frac{1}{n}} \to \frac{\infty}{0} = \infty $
Meaning that $n^n \succ nlog{n}+n\sqrt{n} \hspace{15pt}$(i.e. $n^n$ grows faster than $nlog{n}+n\sqrt{n}$) Notice that if the instead of infinity the limit was zero that would mean that $n^n$ would grow faster than $nlog{n}+n\sqrt{n}$
So, one way to order the functions would be to compute some limits of quotients of functions. However, in general:
$n^n \succ n! \succ c^n \succ n^c \succ log{n} \succ {log(log{n})}$
where c is a constant greater than one. You can actually check this by computing the limits of quotients. by knowing this, it should be easy to order the rest of the functions. Hope I was helpful.