Đáp án:
\[\mathop {\lim }\limits_{x \to 3} \frac{{{x^3} - 5{x^2} + 3x + 9}}{{{x^4} - 8{x^2} - 9}} = 0\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 3} \frac{{{x^3} - 5{x^2} + 3x + 9}}{{{x^4} - 8{x^2} - 9}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x + 1} \right){{\left( {x - 3} \right)}^2}}}{{\left( {{x^2} + 1} \right)\left( {{x^2} - 9} \right)}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x + 1} \right){{\left( {x - 3} \right)}^2}}}{{\left( {{x^2} + 1} \right)\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x + 1} \right)\left( {x - 3} \right)}}{{\left( {{x^2} + 1} \right).\left( {x + 3} \right)}}\\
= \frac{{\left( {3 + 1} \right).\left( {3 - 3} \right)}}{{\left( {{3^2} + 1} \right).\left( {3 + 3} \right)}} = \frac{0}{{60}} = 0
\end{array}\)
