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