Đáp án:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^6} - 3x} }}{{2{x^3} + 1}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{{\sqrt {{x^6} - 3x} }}{{{x^3}}}}}{{\frac{{2{x^3} + 1}}{{{x^3}}}}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {\frac{{{x^6} - 3x}}{{{x^6}}}} }}{{2 + \frac{1}{{{x^3}}}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {1 - \frac{3}{{{x^5}}}} }}{{2 + \frac{1}{{{x^3}}}}}\\
= \frac{{ - 1}}{2}\\
\mathop {\lim }\limits_{x \to + \infty } \left( {x + 2} \right)\sqrt {\frac{{x - 1}}{{{x^3} + 8}}} \\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{{\left( {x + 2} \right)}^2}.\left( {x - 1} \right)} }}{{\sqrt {{x^3} + 8} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{{\left( {1 + \frac{2}{x}} \right)}^2}\left( {1 - \frac{1}{x}} \right)} }}{{\sqrt {1 + \frac{8}{{{x^3}}}} }}\\
= \frac{{1.1}}{1} = 1
\end{array}$
