Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \sqrt {4{x^2} + 3x + 1} + 2x\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{4{x^2} + 3x + 1 - 4{x^2}}}{{\sqrt {4{x^2} + 3x + 1} - 2x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{3x + 1}}{{\sqrt {4{x^2} + 3x + 1} - 2x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{3 + \frac{1}{x}}}{{ - \sqrt {4 + \frac{3}{x} + \frac{1}{{{x^2}}}} - 2}} = \frac{3}{{ - 2 - 2}} = \frac{{ - 3}}{4}
\end{array}\)
