Cho \(f(x)\) là đa thức thỏa mãn \(\mathop {\lim }\limits_{x \to 3} \frac{{f(x) - 15}}{{x - 3}} = 12\).

* Hướng dẫn giải

\(\mathop {\lim }\limits_{x \to 3} \frac{{f(x) - 15}}{{x - 3}} = 12 \Rightarrow f(3) = 15\)

\(T = \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt[3]{{5f(x) - 11}} - 4}}{{{x^2} - x - 6}} = \mathop {\lim }\limits_{x \to 3} \frac{{5(f(x) - 15)}}{{\left( {x - 3} \right)\left( {x + 2} \right)\left( {{{\left( {\sqrt[3]{{5f(x) - 11}}} \right)}^2} + 4\sqrt[3]{{5f(x) - 11}} + 16} \right)}}\)

\(\begin{array}{l}
 = \mathop {\lim }\limits_{x \to 3} \frac{{5.12}}{{\left( {x + 2} \right)\left( {{{\left( {\sqrt[3]{{5f(x) - 11}}} \right)}^2} + 4\sqrt[3]{{5f(x) - 11}} + 16} \right)}}\\
 = \frac{{12}}{{{{\left( {\sqrt[3]{{5f(3) - 11}}} \right)}^2} + 4\sqrt[3]{{5f(3) - 11}} + 16}} = \frac{1}{4}
\end{array}\)