Aloha :)$$a_n=\frac{\sqrt{2n^2+n+2}}{\sqrt{n^2-n-1}+1}=\frac{\frac{1}{n}\sqrt{2n^2+n+2}}{\frac{1}{n}\left(\sqrt{n^2-n-1}+1\right)}=\frac{\sqrt{\frac{1}{n^2}(2n^2+n+2)}}{\sqrt{\frac{1}{n^2}(n^2-n-1)}+\frac{1}{n}}$$$$\phantom{a_n}=\frac{\sqrt{2+\frac{1}{n}+\frac{2}{n^2}}}{\sqrt{1-\frac{1}{n}-\frac{1}{n^2}}+\frac{1}{n}}$$$$\Rightarrow\quad\lim\limits_{n\to\infty}a_n=\frac{\lim\limits_{n\to\infty}\left(\sqrt{2+\frac{1}{n}+\frac{2}{n^2}}\right)}{\lim\limits_{n\to\infty}\left(\sqrt{1-\frac{1}{n}-\frac{1}{n^2}}+\frac{1}{n}\right)}=\frac{\sqrt2}{\sqrt1}=\sqrt2$$
Wegen \(-1\le\cos(n)\le1\) gilt:
$$-1-\frac{1}{n}\le\underbrace{-1+\frac{\cos(n)}{n}}_{=b_n}\le-1+\frac{1}{n}$$$$\lim\limits_{n\to\infty}\left(-1-\frac{1}{n}\right)\le\lim\limits_{n\to\infty}b_n\le\lim\limits_{n\to\infty}\left(-1+\frac{1}{n}\right)$$$$-1\le\lim\limits_{n\to\infty}b_n\le-1$$$$\lim\limits_{n\to\infty}b_n=-1$$
$$c_n=n-\sqrt n+3=\frac{(n-\sqrt n)(n+\sqrt n)}{n+\sqrt n}+3=\frac{n^2-n}{n+\sqrt n}+3$$$$\phantom{c_n}=\frac{n-1}{1+\frac{1}{\sqrt n}}+3\ge\frac{n-1}{1+1}+3=\frac{n-1}{2}+\frac{6}{2}=\frac{n+5}{2}\to\infty$$Die Folge \(c_n\) divergiert.
