Cho hàm số \(f(x)\) xác định trên R và thỏa mãn \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 16}}{{x - 2}} = 12

* Hướng dẫn giải

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{5f\left( x \right) - 16}} - 4}}{{{x^2} + 2x - 8}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {\sqrt[3]{{5f\left( x \right) - 16}} - 4} \right)\left( {{{\sqrt[3]{{5f\left( x \right) - 16}}}^2} + 4\sqrt[3]{{5f\left( x \right) - 16}} + 16} \right)}}{{\left( {x + 4} \right)\left( {x - 2} \right)\left( {{{\sqrt[3]{{5f\left( x \right) - 16}}}^2} + 4\sqrt[3]{{5f\left( x \right) - 16}} + 16} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{5f\left( x \right) - 16 - 64}}{{\left( {x + 4} \right)\left( {x - 2} \right)\left( {{{\sqrt[3]{{5f\left( x \right) - 16}}}^2} + 4\sqrt[3]{{5f\left( x \right) - 16}} + 16} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{5\left[ {f\left( x \right) - 16} \right]}}{{\left( {x + 4} \right)\left( {x - 2} \right)\left( {{{\sqrt[3]{{5f\left( x \right) - 16}}}^2} + 4\sqrt[3]{{5f\left( x \right) - 16}} + 16} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 16}}{{x - 2}}.\frac{5}{{\left( {{{\sqrt[3]{{5f\left( x \right) - 16}}}^2} + 4\sqrt[3]{{5f\left( x \right) - 16}} + 16} \right)}}
\end{array}\)

Ta có \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 16}}{{x - 2}} = 12 \Leftrightarrow f\left( 2 \right) = 16;f'\left( 2 \right) = 12\) 

\( \Rightarrow \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{5f\left( x \right) - 16}} - 4}}{{{x^2} + 2x - 8}} = 12.\frac{5}{{6\left( {{4^2} + 4.4 + 16} \right)}} = \frac{5}{{24}}\)