Uppgift 1

a) Vi använder Maclaurinutveckling:
$$\arctan x=x-\frac13x^3+O(x^5)\ (=x+O(x^3)),$$
vilket ger
$$\begin{aligned} \frac1{(\arctan x)^2}-\frac1{x^2} &=\frac{x^2-(\arctan x)^2}{x^2(\arctan x)^2} =\frac{x^2-(x-\frac13x^3+O(x^5))^2}{x^2(x+O(x^3))^2}\\ &=\frac{\frac23x^4+O(x^6)}{x^4+O(x^6)}=\frac{\frac23+O(x^2)}{1+O(x^2)}\to\frac23. \end{aligned}$$
b) $$\lim_{n\to\infty}\frac{\ln n+\sin n}{\ln n^2+\sin n^2} =\lim_{n\to\infty}\frac{\ln n+\sin n}{2\ln n+\sin n^2} =\lim_{n\to\infty}\frac{1+\frac{\sin n}{\ln n}}{2+\frac{\sin n^2}{\ln n}}=\frac12,$$
eftersom både $\frac{\sin n}{\ln n}$ och $\frac{\sin n^2}{\ln n}$ går mot noll då $n\to\infty$.