Đáp án:
\[\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} - x} + 2x} \right) = \frac{1}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} - x} + 2x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( {\sqrt {4{x^2} - x} + 2x} \right)\left( {\sqrt {4{x^2} - x} - 2x} \right)}}{{\sqrt {4{x^2} - x} - 2x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( {4{x^2} - x} \right) - {{\left( {2x} \right)}^2}}}{{\sqrt {{x^2}\left( {4 - \frac{1}{x}} \right)} - 2x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x}}{{\left| x \right|.\sqrt {4 - \frac{1}{x}} - 2x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x}}{{ - x\sqrt {4 - \frac{1}{x}} - 2x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 1}}{{ - \sqrt {4 - \frac{1}{x}} - 2}}\\
= \frac{{ - 1}}{{ - \sqrt 4 - 2}} = \frac{1}{4}
\end{array}\)
Vậy \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} - x} + 2x} \right) = \frac{1}{4}\)
