Đáp án:
$\begin{array}{l}
5)\mathop {\lim }\limits_{x \to - 1} \frac{{{x^5} + 1}}{{{x^3} + 1}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{\left( {x + 1} \right)\left( {{x^4} - {x^3} + {x^2} - x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{{x^4} - {x^3} + {x^2} - x + 1}}{{{x^2} - x + 1}}\\
= \frac{5}{3}\\
6)\mathop {\lim }\limits_{x \to 3} \frac{{{x^3} - 5{x^2} + 3x + 9}}{{{x^4} - 8{x^2} - 9}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {{x^2} - 2x - 3} \right)}}{{\left( {{x^2} + 1} \right)\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 2x - 3}}{{\left( {{x^2} + 1} \right)\left( {x + 3} \right)}}\\
= \frac{{{3^2} - 2.3 - 3}}{{\left( {{3^2} + 1} \right)\left( {3 + 3} \right)}}\\
= 0
\end{array}$
