Question

Exercițiul 1 și 2. Multumesc

Answer 1

$\\1.) \\ \\ \left(\frac{2x^2+3x}{x-1}\:-\:2x\right) = \frac{2x^2+3x}{x-1}-\frac{2x\left(x-1\right)}{x-1} = \frac{2x^2+3x-2x\left(x-1\right)}{x-1} =\frac{5x}{x-1} \\ \\ => \lim _{x\to \infty \:}\left(\frac{5x}{x-1}\right) => 5\cdot \lim \:_{x\to \infty \:}\left(\frac{x}{x-1}\right) => 5\cdot \lim \:_{x\to \infty \:}\left(\frac{1}{1-\frac{1}{x}}\right) \\ \\=5\cdot \frac{\lim _{x\to \infty \:}\left(1\right)}{\lim _{x\to \infty \:}\left(1-\frac{1}{x}\right)} => 5\cdot \frac{1}{1} = 5$

$\\ 2.) \\ \boxed{Multiplicare \:\ cu \:\ conjugata} \\ \lim _{x\to \infty }\left(\sqrt{x^2-4x+2}-x\right) = \frac{\left(\sqrt{x^2-4x+2}-x\right)\left(\sqrt{x^2-4x+2}+x\right)}{\sqrt{x^2-4x+2}+x} => \\ \\ =\lim _{x\to \infty \:}\left(\frac{-4x+2}{\sqrt{x^2-4x+2}+x}\right) = > \\ \\ \sqrt{x^2-4x+2}+x = \sqrt{x^2\left(1-\frac{4}{x}+\frac{2}{x^2}\right)}+x =>x\sqrt{1-\frac{4}{x}+\frac{2}{x^2}}+x \\ => Acum \:\ vine: \frac{-4x+2}{x\sqrt{1-\frac{4}{x}+\frac{2}{x^2}}+x} =>$

$=\frac{-\frac{4x}{x}+\frac{2}{x}}{\frac{x\sqrt{1-\frac{4}{x}+\frac{2}{x^2}}}{x}+\frac{x}{x}} => \frac{-4+\frac{2}{x}}{\sqrt{1-\frac{4}{x}+\frac{2}{x^2}}+1} \\ \\ \lim _{x\to \infty \:}\left(\frac{-4+\frac{2}{x}}{\sqrt{1-\frac{4}{x}+\frac{2}{x^2}}+1}\right) = \frac{\lim _{x\to \infty \:}\left(-4+\frac{2}{x}\right)}{\lim _{x\to \infty \:}\left(\sqrt{1-\frac{4}{x}+\frac{2}{x^2}}+1\right)} => \\ \\ \lim _{x\to \infty \:}\left(-4+\frac{2}{x}\right) = -4 + 0 = -4$

$\\ \lim _{x\to \infty \:}\left(\sqrt{1-\frac{4}{x}+\frac{2}{x^2}}\right) + \lim _{x\to \infty \:}\left(1\right) \\ \\=> \sqrt{\lim _{x\to \infty \:}\left(1\right)-\lim _{x\to \infty \:}\left(\frac{4}{x}\right)+\lim _{x\to \infty \:}\left(\frac{2}{x^2}\right)} = \sqrt{1-0+0} = 1 \\ => 1+1 = 2 \\ Rezultatul \:\ final: \frac{-4}{2} = \boxed{-2}$