Đáp án:
\[\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4{x^2} - 3x + 4} + 3x}}{{\sqrt {{x^2} + x + 1} - 1}} = - 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4{x^2} - 3x + 4} + 3x}}{{\sqrt {{x^2} + x + 1} - 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left| x \right|\sqrt {4 - \frac{3}{x} + \frac{4}{{{x^2}}}} + 3x}}{{\left| x \right|.\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\sqrt {4 - \frac{3}{x} + \frac{4}{{{x^2}}}} + 3x}}{{ - x.\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - 1}}\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {4 - \frac{3}{x} + \frac{4}{{{x^2}}}} + 3}}{{ - \sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - \frac{1}{x}}}\\
= \frac{{ - \sqrt 4 + 3}}{{ - \sqrt 1 }} = \frac{1}{{ - 1}} = - 1
\end{array}\)
