[tex] \\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 [/tex]
[tex] \\ 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}
=> [/tex]
[tex] =\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
\\ [/tex]
[tex] \\ \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} [/tex]
