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