Prove that $ln(n)^r \in o(n^p)$ for $p>0$ and $r\in \mathbb{R}$ I am trying to proof $f\in o(g)$
Let be $r,p\in \mathbb{R}$ with $p&gt;0$
We have $f(n)=ln^r (n)$ and $g(n)=n^p$
I have already proofed that $ln(n)\in o(n)$ via l'Hospital
$\lim\limits_{n\to \infty}\frac{\frac{1}{x}}{1}=0\Longrightarrow \lim\limits_{n\to \infty}\frac{ln(n)}{n}=0\Longleftrightarrow ln(n)\in o(n)$
I tried substituting n with ln(n) to recieve
$\forall c&gt;0 \exists n_0 \forall n\geq n_0 : ln(ln(n))\leq cln(n)$
Then i tried splitting $c=\frac{a}{b}$ to get $\forall \frac{a}{b}&gt;0\exists n_0 \forall n\geq n_0 : ln(ln(n)) \leq \frac{a}{b}ln(n)$
With this information i said, that $\frac{a}{b}*\frac{b}{a}*b*ln(ln(n))=b*ln(ln(n))\leq \frac{a}{b}*\frac{b}{a}*\frac{a}{b}*b*ln(n)=\frac{a}{b}*b*ln(n)=a*ln(n)$ 
to conclude $ln(n)^b=e^{\frac{a}{b}*\frac{b}{a}*b*ln(ln(n))}\leq e^{\frac{b}{a}*b*ln(n)}=e^{b*ln(n)}=n^a$
At this point i realized a huge problem. I assumed $\frac{a}{b}$ to be positive. But i need it to work for r and p, where only p is known to be postive, which in the other hand means, that r can be negative, what would make $\frac{p}{r}$ negative.
It means that my approach is propably not working a
You just need to show that
$$
\lim_{n \to \infty} \frac{\log^r n}{n^p} = 0.
$$
This is trivial if $r \le 0$ since $\frac{\log^r n}{n^p} = \frac{1}{n^p \log^{-r} n}$ and $\lim_{n \to \infty} n^p \log^{-r} n = +\infty$.
For $r&gt;0$, you can use  l'Hôpital's rule $\lceil r \rceil$ times to obtain:
$$
\lim_{n \to \infty} \frac{\log^r n}{n^p} 
= \lim_{n \to \infty} \frac{r \log^{r-1} n}{pn^{p}} 
= \dots
= \lim_{n \to \infty} \frac{\prod_{i=0}^{\lceil r \rceil-1} (r-i) \cdot \log^{r-\lceil r \rceil} n}{p^{\lceil r \rceil} n^{p}} \\
=
\frac{\prod_{i=0}^{\lceil r \rceil-1} (r-i)}{p^{\lceil r \rceil}}\cdot \lim_{n \to \infty} \frac{ \log^{r-\lceil r \rceil} n}{ n^{p}} =0,
$$
where $\frac{\prod_{i=0}^{\lceil r \rceil-1} (r-i)}{p^{\lceil r \rceil}}$ is a positive constant and
$
\lim_{n \to \infty} \frac{ \log^{r-\lceil r \rceil} n}{ n^{p}}
$
is equal to $0$ since it falls into the previous case.