Giải thích các bước giải:
\(\begin{array}{l}
1,\\
\mathop {\lim }\limits_{x \to - \infty } \frac{{6{x^2} - 5x + 1}}{{1 - 2{x^2}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{6 - \frac{5}{x} + \frac{1}{{{x^2}}}}}{{\frac{1}{{{x^2}}} - 2}} = \frac{{6 - 0 + 0}}{{0 - 2}} = - 3\\
2,\\
\mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {12x - 1} \right)}}{{3{x^3} - 2x + 7}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{12{x^2} - x}}{{3{x^3} - 2x + 7}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{12}}{x} - \frac{1}{{{x^2}}}}}{{3 - \frac{2}{{{x^2}}} + \frac{7}{x}}} = \frac{{0 - 0}}{{3 - 0 + 0}} = 0\\
3,\\
\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {9{x^2} - x + 4} }}{{3x - 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2}\left( {9 - \frac{1}{x} + \frac{4}{{{x^2}}}} \right)} }}{{3x - 1}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{x.\sqrt {9 - \frac{1}{x} + \frac{4}{{{x^2}}}} }}{{3x - 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {9 - \frac{1}{x} + \frac{4}{{{x^2}}}} }}{{3 - \frac{1}{x}}} = \frac{{\sqrt 9 }}{3} = 1\\
4,\\
\mathop {\lim }\limits_{x \to - \infty } \left( {{x^4} - 4{x^2} + 3} \right) = \mathop {\lim }\limits_{x \to - \infty } \left[ {{x^4}.\left( {1 - \frac{4}{{{x^2}}} + \frac{3}{{{x^4}}}} \right)} \right] = + \infty \\
\left( \begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } {x^4} = + \infty \\
\mathop {\lim }\limits_{x \to - \infty } \left( {1 - \frac{4}{{{x^2}}} + \frac{3}{{{x^4}}}} \right) = 1
\end{array} \right)\\
5,\\
\mathop {\lim }\limits_{x \to + \infty } \left( { - 2{x^3} + 3{x^2} + 1} \right) = \mathop {\lim }\limits_{x \to + \infty } \left[ {{x^3}.\left( { - 2 + \frac{3}{x} + \frac{1}{{{x^3}}}} \right)} \right] = - \infty \\
\left( \begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } {x^3} = + \infty \\
\mathop {\lim }\limits_{x \to + \infty } \left( { - 2 + \frac{3}{x} + \frac{1}{{{x^3}}}} \right) = - 2
\end{array} \right)\\
6,\\
\mathop {\lim }\limits_{x \to - \infty } \sqrt {2{x^2} - 3x + 9} = \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^2}.\left( {2 - \frac{3}{x} + \frac{9}{{{x^2}}}} \right)} \\
= \mathop {\lim }\limits_{x \to - \infty } \left( {\left| x \right|.\sqrt {2 - \frac{3}{x} + \frac{9}{{{x^2}}}} } \right) = + \infty \\
\left( \begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \left| x \right| = + \infty \\
\mathop {\lim }\limits_{x \to - \infty } \sqrt {2 - \frac{3}{x} + \frac{9}{{{x^2}}}} = \sqrt 2
\end{array} \right)
\end{array}\)
