Đáp án:
\[\mathop {\lim }\limits_{x \to - \infty } \left( {2x - \sqrt {3{x^2} - x + 1} } \right) = - \infty \]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \left( {2x - \sqrt {3{x^2} - x + 1} } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( {2x - \sqrt {{x^2}.\left( {3 - \frac{1}{x} + \frac{1}{{{x^2}}}} \right)} } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( {2x - \left| x \right|.\sqrt {3 - \frac{1}{x} + \frac{1}{{{x^2}}}} } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( {2x - \left( { - x} \right).\sqrt {3 - \frac{1}{x} + \frac{1}{{{x^2}}}} } \right)\,\,\,\,\,\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \lim \left[ {x.\left( {2 + \sqrt {3 - \frac{1}{x} + \frac{1}{{{x^2}}}} } \right)} \right]\\
= - \infty \\
\left( \begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } x = - \infty \\
\mathop {\lim }\limits_{x \to - \infty } \left( {2 + \sqrt {3 - \frac{1}{x} + \frac{1}{{{x^2}}}} } \right) = 2 + \sqrt 3
\end{array} \right)
\end{array}\)
