I'm looking for some easy way to compare growth rate. I would like to avoid counting the limits of the function.
I heard about a way to compare power's of
$n^{a}(\log n)^{b}$
but i don't know when it does not apply
For exampe, we want to compare growth rate of
$n\over \log n$ and $2^{\sqrt {\log n}}$
I transform it into
$(\log ({n(\log {n})^{-1}}))^1$ and $(\log n)^{1\over 2}$
So is it correct to compare in a way i described before and thereby determine that 1>$1\over 2$ which would mean that $n\over \log n$ grows faster? Or maybe there is a simpler way and I transformed both functions in the wrong way.
You have to be careful, because $\log(\frac{n} {\log n})$ is not of the form $n^a (\log n)^b$, so you cannot make that comparison directly.
If we set $f(n) = \frac{n}{\log n}$ and $g(n) = 2^{\sqrt{\log n}}$, then we can do a more careful comparison by, for example, substituting $N = \log n$ and comparing $\log f(\exp(N))$ to $\log g(\exp(N))$. If we do that, we get
$$\begin{eqnarray} \frac{\log f(\exp(N))}{\log g(\exp(N))} & = & \frac{\log\left(\frac{\exp(N)}{\log \exp(N)}\right)}{\log 2^{\sqrt{\log \exp(N)}}} \\ & = & \frac{\log \exp N - \log \log \exp N}{\sqrt{\log \exp N} / \log 2} \\ & = & \frac{N - \log N}{\sqrt{N} / \log 2} \\ & = & \log 2 \left(\sqrt{N} - \frac{\log N}{\sqrt{N}} \right) \end{eqnarray}$$
From here you can apply what you know - as $N \rightarrow \infty$, $\sqrt{N}$ will also approach infinity, while $\log N$ will grow slower than $\sqrt{N}$ and hence the second term will go to zero. We then work backwards - $N$ going to infinity also means that $n$ is approaching infinity, and so $\log f(n)$ grows faster than $\log g(n)$ and thus $f(n)$ grows faster than $g(n)$.