Aloha :)
Du musst hier prüfen, ob der Grenzwert (Limes Superior) der Quotienten exisitert.
zu a)\(\quad\ln(n)+1=O(\sqrt n)\quad\checkmark\)$$\lim\limits_{n\to\infty}\frac{\ln(n)+1}{\sqrt n}\stackrel{(\ln(n)<\sqrt n)}<\lim\limits_{n\to\infty}\frac{\sqrt n+1}{\sqrt n}=\lim\limits_{n\to\infty}\left(1+\frac{1}{\sqrt n}\right)=1$$
zu b)\(\quad\binom{n}{2}=O(n^2)\quad\checkmark\)$$\lim\limits_{n\to\infty}\frac{\binom{n}{2}}{n^2}=\lim\limits_{n\to\infty}\frac{\frac{n^2-n}{2}}{n^2}=\lim\limits_{n\to\infty}\left(\frac12-\frac{1}{2n}\right)=\frac12$$
zu c)\(\quad2^{n+1}=O(2^n)\quad\checkmark\)$$\lim\limits_{n\to\infty}\frac{2^{n+1}}{2^n}=2$$
zu d)\(\quad3^{2n}=O(3^n)\quad\text{FAIL}\)$$\lim\limits_{n\to\infty}\frac{3^{2n}}{3^n}=\lim\limits_{n\to\infty}\frac{3^n\cdot3^n}{3^n}=\lim\limits_{n\to\infty}3^n=\infty$$
zu e)\(\quad\sqrt n\,\log_2(n)=O(n)\quad\checkmark\)$$\lim\limits_{n\to\infty}\frac{\sqrt n\,\log_2(n)}{n}=\lim\limits_{n\to\infty}\frac{\sqrt n\,\frac{\ln(n)}{\ln(2)}}{n}\stackrel{(\ln(n)<\sqrt n)}<\lim\limits_{n\to\infty}\frac{\frac{n}{\ln(2)}}{n}=\frac{1}{\ln(2)}$$
zu f)\(\quad n^2=O(\ln(n))\quad\text{FAIL}\)$$\lim\limits_{n\to\infty}\frac{n^2}{\ln(n)}>\lim\limits_{n\to\infty}n=\infty$$
