Cho f(x) là đa thức thỏa mãn Lim f(x) - 20 / x - 2 = 10

44/50

Cho \(f\left( x \right)\) là đa thức thỏa mãn \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 20}}{{x - 2}} = 10\). Tính \(T = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{6f\left( x \right) + 5}} - 5}}{{{x^2} + x - 6}}\)

\(T = \frac{{12}}{{25}}\).

\(T = \frac{4}{{25}}\).

\(T = \frac{4}{{15}}\).

\(T = \frac{6}{{25}}\).

Giải thích

Chọn B

Theo giả thiết có \(\mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right) - 20} \right) = 0\)hay \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 20\)\(\left( * \right)\)

Khi đó

\[T = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{6f\left( x \right) + 5}} - 5}}{{{x^2} + x - 6}} = \mathop {\lim }\limits_{x \to 2} \frac{{6f\left( x \right) + 5 - 125}}{{\left( {{x^2} + x - 6} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right) + 25} \right]}}\]

\[T = \mathop {\lim }\limits_{x \to 2} \frac{{6\left[ {f\left( x \right) - 20} \right]}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right) + 25} \right]}}\]

\[T = \frac{{10.6}}{{5.75}} = \frac{4}{{25}}\].