Đáp án:
$\begin{array}{l}
f)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{{\left( {2x + 1} \right)}^3}{{\left( {x + 2} \right)}^{2020}}}}{{\left( {3 - 2{x^4}} \right){{\left( {1 - x} \right)}^{2019}}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\dfrac{{{{\left( {2x + 1} \right)}^3}}}{{{x^3}}}.\dfrac{{{{\left( {x + 2} \right)}^{2020}}}}{{{x^{2020}}}}}}{{\dfrac{{3 - 2{x^4}}}{{{x^4}}}.\dfrac{{{{\left( {1 - x} \right)}^{2019}}}}{{{x^{2019}}}}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{{\left( {2 + \dfrac{1}{x}} \right)}^3}.{{\left( {1 + \dfrac{2}{x}} \right)}^{2020}}}}{{\left( {\dfrac{3}{{{x^4}}} - 2} \right).{{\left( {\dfrac{1}{x} - 1} \right)}^{2019}}}}\\
= \dfrac{{{2^3}{{.1}^{2020}}}}{{3.{{\left( { - 1} \right)}^{2019}}}}\\
= \dfrac{{ - 8}}{3}\\
k)\mathop {\lim }\limits_{x \to + \infty } \dfrac{3}{{\sqrt {4{x^2} + x + 1} - 2x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{3\left( {\sqrt {4{x^2} + x + 1} + 2x} \right)}}{{4{x^2} + x + 1 - 4{x^2}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{3\left( {\sqrt {4{x^2} + x + 1} + 2x} \right)}}{{x + 1}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{3\left( {\sqrt {4 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} + 2} \right)}}{{1 + \dfrac{1}{x}}}\\
= \dfrac{{3.\left( {\sqrt 4 + 2} \right)}}{1}\\
= 12\\
n)\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt[3]{{{x^3} + 2{x^2} + 1}} - 2\sqrt {{x^2} - x} + x} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt[3]{{{x^3} + 2{x^2} + 1}} - x + 2x - 2\sqrt {{x^2} - x} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( \begin{array}{l}
\dfrac{{{x^3} + 2{x^2} + 1 - {x^3}}}{{{{\left( {\sqrt[3]{{{x^3} + 2{x^2} + 1}}} \right)}^2} + x\sqrt[3]{{{x^3} + 2{x^2} + 1}} + {x^2}}}\\
+ \dfrac{{4{x^2} - 4{x^2} + 4x}}{{2x + 2\sqrt {{x^2} - x} }}
\end{array} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( \begin{array}{l}
\dfrac{{2{x^2} + 1}}{{{{\left( {\sqrt[3]{{{x^3} + 2{x^2} + 1}}} \right)}^2} + x\sqrt[3]{{{x^3} + 2{x^2} + 1}} + {x^2}}}\\
+ \dfrac{{4x}}{{2x + 2\sqrt {{x^2} - x} }}
\end{array} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\dfrac{{2 + \dfrac{1}{{{x^2}}}}}{{{{\left( {\sqrt[3]{{1 + \dfrac{2}{x} + \dfrac{1}{{{x^3}}}}}} \right)}^2} + \sqrt[3]{{1 + \dfrac{2}{x} + \dfrac{1}{{{x^3}}}}} + 1}} + \dfrac{4}{{2 + 2\sqrt {1 - \dfrac{1}{x}} }}} \right)\\
= \dfrac{2}{{1 + 1 + 1}} + \dfrac{4}{{2 + 2}}\\
= \dfrac{2}{3} + 1 = \dfrac{5}{3}
\end{array}$
