Uppgift 1

a)
$$\begin{aligned}\lim_{n\to\infty}(\sqrt{n^2+3n}-n)&=\lim_{n\to\infty}\frac{(\sqrt{n^2+3n}-n)(\sqrt{n^2+3n}+n)}{\sqrt{n^2+3n}+n}\\&=\lim_{n\to\infty}\frac{n^2+3n-n^2}{\sqrt{n^2+3n}+n}=\lim_{n\to\infty}\frac{3n}{n(\sqrt{1+3/n}+1)}\\&=\lim_{n\to\infty}\frac3{\sqrt{1+3/n}+1}=\frac3{\sqrt{1+0}+1}=\frac32.\end{aligned}$$
b) Maclaurin-utvecklingen $\sin x=x-\frac16x^3+O(x^5)$ ger
$$\begin{aligned}\sin(\sin x)&=\sin(x-\tfrac16x^3+O(x^5))=(x-\tfrac16x^3+O(x^5))\\&\qquad-\tfrac16(x-\tfrac16x^3+O(x^5))^3+O(x^5)=x-\tfrac13x^3+O(x^5).\end{aligned}$$
Det följer att
$$\begin{aligned}\lim_{x\to0}\frac{\sin(\sin x)-\sin x}{x^3}&=\lim_{x\to0}\frac{x-\frac13x^3+O(x^5)-(x-\frac16x^3+O(x^5))}{x^3}\\&=\lim_{x\to0}\frac{-\frac16x^3+O(x^5)}{x^3}=\lim_{x\to0}(-\tfrac16+O(x^2))=-\frac16.\end{aligned}$$