Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
12,\\
\mathop {\lim }\limits_{x \to - 2} \frac{{{x^3} + 3{x^2} + 2x}}{{{x^2} - x - 6}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{x\left( {x + 1} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x - 3} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{x\left( {x + 1} \right)}}{{x - 3}}\\
= \frac{{\left( { - 2} \right).\left( { - 2 + 1} \right)}}{{ - 2 - 3}}\\
= \frac{2}{{ - 5}} = - \frac{2}{5}\\
13,\\
\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} + 3{x^2} - 9x - 2}}{{{x^3} - x - 6}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {{x^2} + 5x + 1} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 5x + 1}}{{{x^2} + 2x + 3}}\\
= \frac{{{2^2} + 5.2 + 1}}{{{2^2} + 2.2 + 3}}\\
= \frac{{15}}{{11}}
\end{array}\)
