Question

Salut,am nevoie de ajutor la ex. 2

Answer 1

$I_n = \lim\limits_{x\to 0}\dfrac{2x^n - \sin 2x}{x^3}\\ \\ \overset{L'H\, (\frac{0}{0})}{=} \lim\limits_{x\to 0}\dfrac{2nx^{n-1} - 2\cos 2x}{3x^2}\\ \\ \text{Pentru ca limita sa fie tot caz de nedeterminare,}\\\text{singura posibilitate este }\dfrac{2-2}{0}.\\ \\ \Rightarrow 2nx^{n-1} = 2 \Rightarrow n = 1 \Rightarrow p = 1 \\ \\ \Rightarrow I_1 = \lim\limits_{x\to 0}\left(\dfrac{2- 2\cos 2x}{3x^2} \overset{L'H\, (\frac{0}{0})}{=} \dfrac{0+4\sin 2x}{6x} \overset{L'H\, (\frac{0}{0})}{=} \dfrac{8\cos 2x}{6}\right)= \dfrac{4}{3}\\\\ \Rightarrow I_p = \dfrac{4}{3}$

$\\\text{Astfel:}$

$M =\lim\limits_{n\to \infty}\left(1+\dfrac{1}{n}I_{p}\right)^n=\lim\limits_{n\to \infty}\left(1+\dfrac{1}{n}\cdot \dfrac{4}{3}\right)^n= \\ \\ =\left[\lim\limits_{n\to \infty} \left(1+\dfrac{1}{\dfrac{3}{4}n}\right)^{\dfrac{3}{4}n}\right]^{\dfrac{4}{3}} =e^{\dfrac{4}{3}}= \boxed{e\sqrt[3]{e}}\\ \\ \\\Rightarrow \textbf{a) corect}$