Giải thích các bước giải:
$\begin{array}{l}
a)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3x + 1} \right) + {x^3} - x}}{{{x^3} + 2x + 1}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^3} + 2x + 1}}{{{x^3} + 2x + 1}}\\
= \mathop {\lim }\limits_{x \to + \infty } 1\\
= 1\\
b)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3x + 2} \right){{\left( {x - 1} \right)}^2}}}{{\left( {5x - 3} \right){{\left( {2x + 3} \right)}^3}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\dfrac{{\left( {3x + 2} \right){{\left( {x - 1} \right)}^2}}}{{{x^4}}}}}{{\dfrac{{\left( {5x - 3} \right){{\left( {2x + 3} \right)}^3}}}{{{x^4}}}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {\dfrac{3}{x} + \dfrac{2}{{{x^2}}}} \right){{\left( {1 - \dfrac{1}{x}} \right)}^2}}}{{\left( {5 - \dfrac{3}{x}} \right){{\left( {2 + \dfrac{3}{x}} \right)}^3}}}\\
= \dfrac{{\left( {3.0 + 2.0} \right){{\left( {1 - 0} \right)}^2}}}{{\left( {5 - 3.0} \right){{\left( {2 + 3.0} \right)}^3}}}\\
= 0
\end{array}$
