Uppgift 1

a) Vi förlänger med konjugatuttrycket:
$$\begin{aligned} \lim_{n\to\infty}(\sqrt{n^2+3n+5}-n) &=\lim_{n\to\infty}\frac{(\sqrt{n^2+3n+5}-n)(\sqrt{n^2+3n+5}+n)}{\sqrt{n^2+3n+5}+n}\\ &=\lim_{n\to\infty}\frac{(n^2+3n+5)-n^2}{\sqrt{n^2+3n+5}+n} =\lim_{n\to\infty}\frac{3n+5}{n\sqrt{1+3/n+5/n^2}+n}\\ &=\lim_{n\to\infty}\frac{3+5/n}{\sqrt{1+3/n+5/n^2}+1}=\frac{3+0}{1+1}=\frac32. \end{aligned}$$
b) Vi använder Maclaurinutveckling:
$$\arctan x=x-\frac13x^3+O(x^5)$$
vilket ger $\arctan2x-2\arctan x=$
$$(2x-\tfrac83x^3+O(x^5))-2(x-\tfrac13x^3+O(x^5))=-2x^3+O(x^5).$$
På liknande sätt ger $e^t=1+t+O(t^2)$:
$$e^{x^3}-1=x^3+O(x^6).$$
Tillsammans ger detta
$$\frac{\arctan2x-2\arctan x}{e^{x^3}-1}=\frac{-2x^3+O(x^5)}{x^3+O(x^6)}=\frac{-2+O(x^2)}{1+O(x^3)}\to-2.$$