Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {9{x^2} + x + 1} + 3x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {\sqrt {9{x^2} + x + 1} + 3x} \right)\left( {\sqrt {9{x^2} + x + 1} - 3x} \right)}}{{\sqrt {9{x^2} + x + 1} - 3x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {9{x^2} + x + 1} \right) - {{\left( {3x} \right)}^2}}}{{\sqrt {{x^2}\left( {9 + \frac{1}{x} + \frac{1}{{{x^2}}}} \right)} - 3x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x + 1}}{{\left| x \right|.\sqrt {9 + \frac{1}{x} + \frac{1}{{{x^2}}}} - 3x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x + 1}}{{ - x\sqrt {9 + \frac{1}{x} + \frac{1}{{{x^2}}}} - 3x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{1 + \frac{1}{x}}}{{ - \sqrt {9 + \frac{1}{x} + \frac{1}{{{x^2}}}} - 3}}\\
= \dfrac{1}{{ - \sqrt 9 - 3}} = - \dfrac{1}{6}
\end{array}\)
