Aloha :)
zu (a)$$a_n=\frac{\sqrt{2n^2+n+2}}{\sqrt{n^2-n-1}+1}=\frac{\frac{1}{n}\sqrt{2n^2+n+2}}{\frac{1}{n}\left(\sqrt{n^2-n-1}+1\right)}$$$$\phantom{a_n}=\frac{\sqrt{\frac{1}{n^2}(2n^2+n+2)}}{\sqrt{\frac{1}{n^2}(n^2-n-1)}+\frac{1}{n}}=\frac{\sqrt{2+\frac{1}{n}+\frac{2}{n^2}}}{\sqrt{1-\frac{1}{n}-\frac{1}{n^2}}+\frac{1}{n}}$$$$\Rightarrow\quad\lim\limits_{n\to\infty}a_n=\frac{\lim\limits_{n\to\infty}\left(\sqrt{2+\frac{1}{n}+\frac{2}{n^2}}\right)}{\lim\limits_{n\to\infty}\left(\sqrt{1-\frac{1}{n}-\frac{1}{n^2}}+\frac{1}{n}\right)}=\frac{\sqrt2}{\sqrt1}=\sqrt2$$
zu (b)
Wegen \(-1\le\cos(n)\le1\) gilt:$$\phantom{\Rightarrow\quad}-2-\frac{1}{n}\le\underbrace{-2+\frac{\cos(n)}{n}}_{=b_n}\le-2+\frac{1}{n}$$$$\Rightarrow\quad\lim\limits_{n\to\infty}\left(-2-\frac{1}{n}\right)\le\lim\limits_{n\to\infty}b_n\le\lim\limits_{n\to\infty}\left(-2+\frac{1}{n}\right)$$$$\Rightarrow\quad-2\le\lim\limits_{n\to\infty}b_n\le-2$$$$\Rightarrow\quad\lim\limits_{n\to\infty}b_n=-2$$
zu (c)$$c_n=n-\sqrt n+5=\frac{(n-\sqrt n)(n+\sqrt n)}{n+\sqrt n}+5=\frac{n^2-n}{n+\sqrt n}+5$$$$\phantom{c_n}=\frac{n-1}{1+\frac{1}{\sqrt n}}+5\ge\frac{n-1}{1+1}+5=\frac{n-1}{2}+\frac{10}{2}=\frac{n+9}{2}\to\infty$$Die Folge \(c_n\) divergiert.
zu (d)$$a_n=\frac{5^n-3^n}{2^n+5^{n+1}+5^{n+2}}=\frac{\frac{5^n}{5^n}-\frac{3^n}{5^n}}{\frac{2^n}{5^n}+\frac{5^{n+1}}{5^n}+\frac{5^{n+2}}{5^n}}=\frac{1-\left(\frac{3}{5}\right)^n}{\left(\frac{2}{5}\right)^n+5^1+5^2}=\frac{1-\left(\frac{3}{5}\right)^n}{30+\left(\frac{2}{5}\right)^n}$$$$\lim\limits_{n\to\infty}a_n=\frac{\lim\limits_{n\to\infty}\left(1-\left(\frac{3}{5}\right)^n\right)}{\lim\limits_{n\to\infty}\left(30+\left(\frac{2}{5}\right)^n\right)}=\frac{1}{30}$$
