Giải thích các bước giải:
\(begin{array}{l}
1,\\
\mathop {\lim }\limits_{x \to 2} \frac{{3{x^3} - 5{x^2} - 3x + 2}}{{ - {x^3} + 3{x^2} - x - 2}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\left( {3{x^3} - 6{x^2}} \right) + \left( {{x^2} - 2x} \right) - \left( {x - 2} \right)}}{{\left( { - {x^3} + 2{x^2}} \right) + \left( {{x^2} - 2x} \right) + \left( {x - 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {3{x^2} + x - 1} \right)}}{{\left( {x - 2} \right)\left( { - {x^2} + x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} + x - 1}}{{ - {x^2} + x + 1}}\\
= \frac{{{{3.2}^2} + 2 - 1}}{{ - {2^2} + 2 + 1}}\\
= - 13\\
3,\\
\mathop {\lim }\limits_{x \to - \frac{1}{2}} \frac{{2{x^3} + {x^2} + 6x + 3}}{{4{x^3} + 14{x^2} + 14x + 4}}\\
= \mathop {\lim }\limits_{x \to - \frac{1}{2}} \frac{{\left( {2x + 1} \right)\left( {{x^2} + 3} \right)}}{{2\left( {2x + 1} \right)\left( {x + 1} \right)\left( {x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - \frac{1}{2}} \frac{{{x^2} + 3}}{{2\left( {x + 1} \right)\left( {x + 2} \right)}}\\
= \frac{{{{\left( { - \frac{1}{2}} \right)}^2} + 3}}{{2.\left( { - \frac{1}{2} + 1} \right)\left( { - \frac{1}{2} + 2} \right)}} = \frac{{13}}{6}\\
5,\\
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{{x^2} + x + 1}} - \sqrt[3]{{{x^3} + 1}}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {{x^2} + x + 1} \right) - \left( {{x^3} + 1} \right)}}{{\left( {\sqrt[3]{{{{\left( {{x^2} + x + 1} \right)}^2}}} + \sqrt[3]{{{x^2} + x + 1}}.\sqrt[3]{{{x^3} + 1}} + \sqrt[3]{{{{\left( {{x^3} + 1} \right)}^2}}}} \right).x}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{ - {x^2} + x + 1}}{{\sqrt[3]{{{{\left( {{x^2} + x + 1} \right)}^2}}} + \sqrt[3]{{{x^2} + x + 1}}.\sqrt[3]{{{x^3} + 1}} + \sqrt[3]{{{{\left( {{x^3} + 1} \right)}^2}}}}}\\
= \frac{1}{{1 + 1 + 1}} = \frac{1}{3}
\end{array}\)
