Question

O limita când x tinde la infinit

Answer 1

$\lim\limits_{x\to \infty}\left[x^2\Big(\arctan(x+1) - \arctan x\Big)\right] = \\ \\ = \lim\limits_{x\to \infty}\left[x^2\cdot \arctan \Big(\dfrac{x+1-x}{1+(x+1)x}\Big)\right] = \\ \\ =\lim\limits_{x\to \infty}\left[x^2\cdot \arctan \Big(\dfrac{1}{x^2+x+1}\Big)\right]= \\ \\ \\= \lim\limits_{x\to \infty}\left[\dfrac{\arctan \Big(\dfrac{1}{x^2+x+1}\Big)}{\dfrac{1}{x^2+x+1}}\cdot \dfrac{x^2}{x^2+x+1}\right] = \\ \\ \\=\lim\limits_{x\to \infty}\left[1\cdot \dfrac{x^2}{x^2+x+1}\right] = 1$

$\boxed{\lim\limits_{u(x) \to 0}\dfrac{\arctan\Big(u(x)\Big)}{u(x)} = 1}\rightarrow \text{limita remarcabila}$