Đáp án: `Lim_{x-> +\infty}`$\dfrac{2x^2-x.\sqrt{x^2-4}}{x^2-3x+2}=1$
Giải thích các bước giải:
`Lim_{x-> +\infty}`$\dfrac{2x^2-x.\sqrt{x^2-4}}{x^2-3x+2}=$`Lim_{x-> +\infty}`$\dfrac{2x^2-x.x.\sqrt{ \dfrac{x^2}{x^2}- \dfrac{4}{x^2}}}{x^2-3x+2}$
`=Lim_{x-> +\infty}`$\dfrac{ \dfrac{2x^2}{x^2}- \dfrac{x^2}{x^2}.\sqrt{ 1- \dfrac{4}{x^2}}}{ \dfrac{x^2}{x^2}- \dfrac{3x}{x^2}+ \dfrac{2}{x^2}}$ `=Lim_{x-> +\infty}`$\dfrac{ 2- 1.\sqrt{ 1- \dfrac{4}{x^2}}}{ 1- \dfrac{3}{x}+ \dfrac{2}{x^2}}$
$=\dfrac{ 2- \sqrt{ 1- 0}}{ 1- 0+ 0}=\dfrac{2-1}{1}=1$
