Cho \(f\left( x \right)\) là đa thức thỏa mãn \(\mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 27}}{{x - 2}} = 5\). Tính \(I = \mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt[3]{{2f\left( x \right) + 10}} - 4}}{{{x^2} + x - 6}}\)

Chỉ điền số nguyên, phân số dạng a/b

Đáp án:

Đặt \(g\left( x \right) = \dfrac{{f\left( x \right) - 27}}{{x - 2}}\)

\(\begin{array}{l} \Rightarrow f\left( x \right) = 27 + \left( {x - 2} \right)g\left( x \right)\\ \Rightarrow \mathop {\lim }\limits_{x \to 3} f\left( x \right) = 27\end{array}\)

\(\begin{array}{l}I = \mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt[3]{{2f\left( x \right) + 10}} - 4}}{{{x^2} + x - 6}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{2f\left( x \right) + 10 - 64}}{{\left( {{x^2} + x - 6} \right)\left[ {{{\left( {\sqrt[3]{{2f\left( x \right) + 10}}} \right)}^2} + 4\sqrt[3]{{2f\left( x \right) + 10}} + 16} \right]}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{2f\left( x \right) + 10 - 64}}{{\left( {{x^2} + x - 6} \right)\left[ {{{\left( {\sqrt[3]{{2f\left( x \right) + 10}}} \right)}^2} + 4\sqrt[3]{{2f\left( x \right) + 10}} + 16} \right]}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{2f\left( x \right) - 54}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{2f\left( x \right) + 10}}} \right)}^2} + 4\sqrt[3]{{2f\left( x \right) + 10}} + 16} \right]}}\\ = \mathop {\lim }\limits_{x \to 2} \left[ {\dfrac{{f\left( x \right) - 27}}{{x - 2}}.\dfrac{2}{{\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{2f\left( x \right) + 10}}} \right)}^2} + 4\sqrt[3]{{2f\left( x \right) + 10}} + 16} \right]}}} \right]\\ = 5.\dfrac{2}{{\left( {2 + 3} \right)48}} = \dfrac{1}{{24}}\end{array}\)