Đáp án:
$\begin{array}{l}
a)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^3} + 5{x^2} - 3}}{{{x^2} + 6x + 3}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x + 5 - \dfrac{3}{{{x^2}}}}}{{1 + \dfrac{6}{x} + \dfrac{3}{{{x^2}}}}}\\
= - \infty \\
b)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^3} - 7{x^2} + 11}}{{3{x^6} + 2{x^5} - 5}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\dfrac{2}{{{x^3}}} - \dfrac{7}{{{x^4}}} + \dfrac{{11}}{{{x^6}}}}}{{3 + \dfrac{2}{x} - \dfrac{5}{{{x^6}}}}}\\
= 0\\
c)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x - 3}}{{\sqrt {{x^2} + 1} - x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{2 - \dfrac{3}{x}}}{{ - \sqrt {1 + \dfrac{1}{{{x^2}}}} - 1}}\\
= \dfrac{2}{{ - 1 - 1}}\\
= - 1\\
d)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {4{x^2} - x + 1} }}{{x + 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - \sqrt {4 - \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} }}{{1 + \dfrac{1}{x}}}\\
= \dfrac{{ - \sqrt 4 }}{1}\\
= - 2\\
e)\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt[3]{{3{x^3} - 1}} + \sqrt {{x^2} + 2} } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } x.\left( {\sqrt[3]{{3 - \dfrac{1}{{{x^3}}}}} - \sqrt {1 + \dfrac{2}{{{x^2}}}} } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } x.\left( {\sqrt[3]{3} - 1} \right)\\
= - \infty
\end{array}$
