Question

Buna !! Multumesc!!..............

Answer 1

$\textbf{Limita remarcabila:}\\ \\ \lim\limits_{x\to 0}\dfrac{\arctan x}{x} = 1$

$\textbf{Rezolvare:}\\ \\\lim\limits_{x\to 0}\dfrac{\arctan^2 x}{\cos 2x - e^{x^2}} =\lim\limits_{x\to 0}\left[\left(\dfrac{\arctan x}{x}\right)^2\cdot \dfrac{x^2}{\cos 2x-e^{x^2}}\right] = \\ \\ =1^2\cdot \lim\limits_{x\to 0}\dfrac{x^2}{\cos 2x - e^{x^2}} \overset{(L'H.)}{=}\lim\limits_{x\to 0}\dfrac{2x}{-2\sin 2x - 2xe^{x^2}}\overset{(L'H.)}{=} \\ \\ \overset{(L'H.)}{=} \lim\limits_{x\to 0}\dfrac{2}{-4\cos 2x - 2e^{x^2}-4x^2e^{x^2}}=\\ \\ =\dfrac{2}{-4\cdot 1-2\cdot 1-0} = -\dfrac{2}{6} = \boxed{-\dfrac{1}{3}}$