Uppgift 1

a) $$\begin{aligned}\lim_{n\to\infty}(\sqrt{n^2+3n}-n)&=\lim_{n\to\infty}\frac{(\sqrt{n^2+3n}-n)(\sqrt{n^2+3n}+n)}{\sqrt{n^2+3n}+n}\\&=\lim_{n\to\infty}\frac{n^2+3n-n^2}{\sqrt{n^2+3n}+n}=\lim_{n\to\infty}\frac{3n}{\sqrt{n^2+3n}+n}\\&=\lim_{n\to\infty}\frac3{\sqrt{1+3/n}+1}=\frac3{1+1}=\frac32.\end{aligned}$$
b) Maclaurinutveckling ger
$$2\arctan x=2(x-\tfrac13x^3+O(x^5))=2x-\tfrac23x^3+O(x^5),$$
$$\arctan2x=2x-\tfrac13(2x)^3+O(x^5)=2x-\tfrac83x^3+O(x^5),$$
$$3\sin x=3(x-\tfrac16x^3+O(x^5))=3x-\tfrac12x^3+O(x^5),$$
$$\sin3x=3x-\tfrac16(3x)^3+O(x^5)=3x-\tfrac92x^3+O(x^5).$$
Det följer att
$$\begin{aligned}\lim_{x\to0}\frac{2\arctan x-\arctan2x}{3\sin x-\sin3x}&=\lim_{x\to0}\frac{(2x-\frac23x^3+O(x^5))-(2x-\frac83x^3+O(x^5))}{(3x-\frac12x^3+O(x^5))-(3x-\frac92x^3+O(x^5))}\\&=\lim_{x\to0}\frac{2x^3+O(x^5)}{4x^3+O(x^5)}=\lim_{x\to0}\frac{2+O(x^2)}{4+O(x^2)}=\frac12.\end{aligned}$$