tìm lim ( √x ² +x +2x) x-> - ∞ A. 2 B. - ∞ C. 1 D. + ∞ Giải chi tiết giúp em

Đáp án:

$B.$ $-\infty \:$

Giải thích các bước giải:

$\lim _{x\to -\infty }\left(\sqrt{x^2+x}+2x\right)$

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

$=\lim _{x\to \:-\infty \:}\left(1+\dfrac{2x}{\sqrt{x^2+x}}\right). \lim _{x\to \:-\infty \:}\left(\sqrt{x^2+x}\right)$

$=\left(-1\right). \infty \:$

$=-\infty \:$

$\begin{array}{l} \mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {{x^2} + x}  + 2x} \right)\\  = \mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {{x^2}\left( {1 + \dfrac{1}{x}} \right)}  + 2x} \right)\\  = \mathop {\lim }\limits_{x \to  - \infty } \left( {\left| x \right|\sqrt {1 + \dfrac{1}{x}}  + 2x} \right)\\  = \mathop {\lim }\limits_{x \to  - \infty } \left( { - x\sqrt {1 + \dfrac{1}{x}}  + 2x} \right) = \mathop {\lim }\limits_{x \to  - \infty } x\left( {2 - \sqrt {1 + \dfrac{1}{x}} } \right)\\  =  - \infty  \to B \end{array}$