Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^3} - 3{x^2} + 5x - 6}}{{1 - 3{x^3}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{1 - \frac{3}{x} + \frac{5}{{{x^2}}} - \frac{6}{{{x^3}}}}}{{\frac{1}{{{x^3}}} - 3}}\\
= \frac{{1 - 0 + 0 - 0}}{{0 - 3}}\\
= - \frac{1}{3}\\
b,\\
\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {2{x^2} + 3x - 5} }}{{3x + 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2}.\left( {2 + \frac{3}{x} - \frac{5}{{{x^2}}}} \right)} }}{{3x + 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left| x \right|.\sqrt {2 + \frac{3}{x} - \frac{5}{{{x^2}}}} }}{{3x + 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( { - x} \right).\sqrt {2 + \frac{3}{x} - \frac{5}{{{x^2}}}} }}{{3x + 1}}\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {2 + \frac{3}{x} - \frac{5}{{{x^2}}}} }}{{3 + \frac{1}{x}}}\\
= \frac{{ - \sqrt {2 + 0 - 0} }}{{3 + 0}}\\
= \frac{{ - \sqrt 2 }}{3}
\end{array}\)
