limit $n^{1/r}$ and $n^r$ I apologize in advance for this super basic questions,
but how to proof that
$\lim_{n \rightarrow \infty} \sqrt[r]{{n}}=\infty=\lim_{n \rightarrow \infty} n^{r}$ (for r>0) do I have to show it with $\epsilon-\delta$ or can it be argued faster?
 A: For any value $M > 0$ show that you can find $N$ such that $f(n) \ge M$ for all $n \ge N$.
For example, with $f(n) = n^r$, note that $N=\lceil M^{1/r}\rceil$ will lead to $n^r \ge N^r \ge M$ for all $n \ge N$.
A: It's most straightforward to use an $M-N$ argument.
But f we accept that $\sqrt[r]{x}$ and $x^r$ are continuous then we can do a "limit of a function is the function of the limit" argument: $\lim_{n\to \infty} \sqrt[r]{n} = \sqrt[r]{\lim_{n\to \infty} n}$ and $\lim_{n\to\infty}n^r = (\lim_{n\to\infty} n)^r$.
But I'd advise against it under the "if you have to ask, you need more practice" rule. It encourages looking for fast solutions without really attempting to understand and does some sloppy "treat $\infty$ as though it were a number" abuses.
And if one can apply rules like a monkey grinding an organ, then one can apply and $M-N$ argument.
For any $N\ge 0$ let $M =N^r$.  If $n > M$ then $n\ge N^r$ and $\sqrt[r]n > N$.  SO $\lim_{n\to \infty}  \sqrt[r] n =\infty$.
Same thing for $M' = \sqrt[r]N$.  If $n > M'$ then $n \ge \sqrt[r]N$ and $n^r \ge N$. So $\lim_{n\to\infty} n^r = \infty$.
