tìm lim ( √x ²+x+2 + x +2 x->- ∞ A 3/2 B 0 C - ∞ D -2 giải chi tiết giúp em ạ

Đáp án:

`A` 

Giải thích các bước giải:

$\lim\limits_{x \to -∞}$`\sqrt{x^{2}+x+2}+x+2`

$=\lim\limits_{x \to -∞}$`\frac{x^{2}+x+2-(x+2)^{2}}{\sqrt{x^{2}+x+2}-(x+2)}`

$=\lim\limits_{x \to -∞}$`\frac{x^{2}+x+2-(x^{2}+4x+4)}{\sqrt{x^{2}+x+2}-(x+2)}`

$=\lim\limits_{x \to -∞}$`\frac{x^{2}+x+2-x^{2}-4x-4){\sqrt{x^{2}+x+2}-(x+2)}`

$=\lim\limits_{x \to -∞}$`\frac{-3x-2}{\sqrt{x^{2}+x+2}-x-2}`

$=\lim\limits_{x \to -∞}$`\frac{x.(-3-\frac{2}{x})}{|x|\sqrt{1+\frac{1}{x}+\frac{2}{x^{2}}}-x-2}`

$=\lim\limits_{x \to -∞}$`\frac{x.(-3-\frac{2}{x})}{x(-\sqrt{1+\frac{1}{x}+\frac{2}{x^{2}}}-1-\frac{2}{x})`

$=\lim\limits_{x \to -∞}$`\frac{-3-\frac{2}{x}}{-\sqrt{1+\frac{1}{x}+\frac{2}{x^{2}}}-1-\frac{2}{x}`

`=\frac{-3}{-2}=\frac{3}{2}`