Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\mathop {\lim }\limits_{x \to - 2} \left( {3{x^2} - 3x - 8} \right) = 3.{\left( { - 2} \right)^2} - 3.\left( { - 2} \right) - 8 = 10\\
b,\\
\mathop {\lim }\limits_{x \to - 1} \frac{{3{x^3} - {x^2} + 2}}{{x - 2}} = \frac{{3.{{\left( { - 1} \right)}^3} - {{\left( { - 1} \right)}^2} + 2}}{{\left( { - 1} \right) - 2}} = \frac{{ - 2}}{{ - 3}} = \frac{2}{3}\\
c,\\
\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 3x + 2}}{{2x - 4}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{2\left( {x - 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{x - 1}}{2} = \frac{{2 - 1}}{2} = \frac{1}{2}\\
d,\\
\mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - {x^3}}}{{{x^2} - x + 3}} = \frac{{{{\left( { - 2} \right)}^2} - {{\left( { - 2} \right)}^3}}}{{{{\left( { - 2} \right)}^2} - \left( { - 2} \right) + 3}} = \frac{{12}}{9} = \frac{4}{3}
\end{array}\)
