I am looking for a definition of "which function grows faster", I came across this but I'm not sure if it's right:
$\rightarrow$ $e^x >> x^k >> \ln(x)$ (for $x \rightarrow \infty$ and $k \in \mathbb{R}$)
Is this right? Or should it be $k > 1$? So the main question is, if $x^{0.00001}$ still grows faster than $\ln(x)$. And the second question is: shouldn't it be only positive real numbers? because $x^{-1}$ with $x$ going to infinity goes to $0$, right?
Let $$f(x) = x^{1/k}, \\ g(x) = \log x.$$ Then $$f(e^x) = e^{x/k}, \\ g(e^x) = x.$$ So it becomes obvious that if $x$ is sufficiently large, $f > g$. For instance, if $k = 10000$ as in your case, we have $$f(e^{10^6}) = e^{10^6/10^4} = e^{100} > 2^{100} = 16^{25} > 10^{25} \gg 10^6 = g(e^{10^6}).$$
Yes, you're right that $\lim_{x\to+\infty} {\log x\over x^h}=0$ only for $h>0$. But/and it is not necessary that $h>1$. If you graphically compare $x^{.000001}$ and $\log x$, you will be deceived about the eventual behavior. :)
The standard proof that $e^x\gg x^k$ for all $k=1,2,\ldots$ (and $x\to+\infty$) is $$ e^x \;\gg\; {x^{k+1}\over (k+1)!} \;\gg\; x^k $$ for $x>(k+1)!$, etc. Taking logarithms, (since $\log$ is monotone increasing) $x\gg k\log x$. Replacing $x$ by $x^{1/k}$, $x^{1/k}\gg \log x$ for $k=1,2,3,\ldots$ This suffices to give $x^h\gg \log x$ for any positive $h$.
hint
Let us compute $$L=\lim_{x\to+\infty}\frac{\ln(x)}{x^k}$$
If $ k=0 $, then $ L=+\infty$.
If $ k<0 $, then $$L=\lim_{x\to+\infty}x^{-k}\ln(x)=+\infty$$
If $ k>0 $, then
$$L=\lim_{x\to+\infty}\frac 1k\frac{\ln(x^k)}{x^k}=0$$
So, $$x^k>>\ln(x) (x\to+\infty) \;\iff\; k>0$$
This would seem to answer everything:
