Giải thích các bước giải:
\(\begin{array}{l}
7,\\
\mathop {\lim }\limits_{x \to 3} \frac{{3{x^3} - 7x - 60}}{{9 - {x^2}}} = \mathop {\lim }\limits_{x \to 3} \frac{{3{x^3} - 9{x^2} + 9{x^2} - 27x + 20x - 60}}{{\left( {3 - x} \right)\left( {3 + x} \right)}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {3{x^2} + 9x + 20} \right)}}{{\left( {3 - x} \right)\left( {3 + x} \right)}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{3{x^2} + 9x + 20}}{{ - \left( {3 + x} \right)}}\\
= \frac{{{{3.3}^2} + 9.3 + 20}}{{ - \left( {3 + 3} \right)}} = - \frac{{37}}{3}\\
8,\\
\mathop {\lim }\limits_{x \to - 3} \frac{{{x^4} - 10{x^2} + 9}}{{x + 3}}\\
= \mathop {\lim }\limits_{x \to - 3} \frac{{\left( {{x^2} - 1} \right)\left( {{x^2} - 9} \right)}}{{x + 3}}\\
= \mathop {\lim }\limits_{x \to - 3} \left[ {\left( {{x^2} - 1} \right)\left( {x - 3} \right)} \right]\\
= \left( {9 - 1} \right)\left( { - 3 - 3} \right) = - 48
\end{array}\)
