Đáp án:
$\begin{array}{l}
a)\mathop {\lim }\limits_{x \to \infty } \dfrac{5}{{3x + 2}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\dfrac{5}{x}}}{{3 + \dfrac{2}{x}}} = 0\\
b)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^4} + 7}}{{{x^4} + 1}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{1 + \dfrac{7}{{{x^4}}}}}{{1 + \dfrac{1}{{{x^4}}}}} = 1\\
c)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x - \sqrt {3{x^2} + 2} }}{{5x + \sqrt {{x^2} + 1} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{2 - \sqrt {3 + \dfrac{2}{{{x^2}}}} }}{{5 + \sqrt {1 + \dfrac{1}{{{x^2}}}} }}\\
= \dfrac{{2 - \sqrt 3 }}{{5 + \sqrt 1 }} = \dfrac{{2 - \sqrt 3 }}{6}\\
d)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{{\left( {2x - 3} \right)}^{20}}.{{\left( {3x + 2} \right)}^{30}}}}{{{{\left( {2x + 1} \right)}^{50}}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{{\left( {\dfrac{{2x - 3}}{x}} \right)}^{20}}.{{\left( {\dfrac{{3x + 2}}{x}} \right)}^{30}}}}{{{{\left( {\dfrac{{2x + 1}}{x}} \right)}^{50}}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{{\left( {2 - \dfrac{3}{x}} \right)}^{20}}.{{\left( {3 + \dfrac{2}{x}} \right)}^{30}}}}{{{{\left( {2 + \dfrac{1}{x}} \right)}^{50}}}}\\
= \dfrac{{{2^{20}}{{.3}^{30}}}}{{{2^{50}}}}\\
= {\left( {\dfrac{3}{2}} \right)^{30}}\\
e)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{x\sqrt x + 1}}{{{x^2} + x + 1}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\dfrac{1}{{\sqrt x }} + \dfrac{1}{{{x^2}}}}}{{1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}}} = 0\\
f)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {2{x^4} + {x^2} - 1} }}{{1 - 2x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {2{x^2} + 1 - \dfrac{1}{{{x^2}}}} }}{{\dfrac{1}{x} - 2}}\\
= + \infty
\end{array}$
