tìm lim ( √x ² +x +2x) x-> - ∞ A. 2 B. - ∞ C. 1 D. + ∞ Giải chi tiết giúp em
2 câu trả lời
Đáp án:
$B.$ $-\infty \:$
Giải thích các bước giải:
$\lim _{x\to -\infty }\left(\sqrt{x^2+x}+2x\right)$
$=\lim _{x\to \:-\infty \:}\left(\left(1+\dfrac{2x}{\sqrt{x^2+x}}\right)\sqrt{x^2+x}\right)$
$=\lim _{x\to \:-\infty \:}\left(1+\dfrac{2x}{\sqrt{x^2+x}}\right). \lim _{x\to \:-\infty \:}\left(\sqrt{x^2+x}\right)$
$=\left(-1\right). \infty \:$
$=-\infty \:$
$\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + x} + 2x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2}\left( {1 + \dfrac{1}{x}} \right)} + 2x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( {\left| x \right|\sqrt {1 + \dfrac{1}{x}} + 2x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( { - x\sqrt {1 + \dfrac{1}{x}} + 2x} \right) = \mathop {\lim }\limits_{x \to - \infty } x\left( {2 - \sqrt {1 + \dfrac{1}{x}} } \right)\\
= - \infty \to B
\end{array}$