Cho hàm số \(f\left( x \right)\) xác định trên \(\mathbb{R}\) thỏa mãn \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 16}} = 12.\) Giới hạn \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {2f\left( x \right) - 16} - 4}}{{{x^2} + x - 6}}\) bằng

Ta có \(f\left( x \right) = \left( {x - 2} \right)g\left( x \right) + 16\)\( \Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left[ {\left( {x - 2} \right)g\left( x \right) + 16} \right] = 16.{\rm{ }}\)

Khi đó \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {2f\left( x \right) - 16}  - 4}}{{{x^2} + x - 6}}\)\[ = \mathop {\lim }\limits_{x \to 2} \frac{{2f\left( x \right) - 16 - 16}}{{\left( {{x^2} + x - 6} \right)\left( {\sqrt {2f\left( x \right) - 16}  + 4} \right)}}\]\[g\left( x \right) = \frac{{f\left( x \right) - 16}}.\]

\( = \mathop {\lim }\limits_{x \to 2} \frac{{2f\left( x \right) - 32}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left( {\sqrt {2f\left( x \right) - 16}  + 4} \right)}}\)\( = \mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 16}} \cdot \mathop {\lim }\limits_{x \to 2} \frac{2}{{\left( {x + 3} \right)\left( {\sqrt {2f\left( x \right) - 16}  + 4} \right)}}\)

\( = 12.\frac{2} = \frac{3}{5}.\) Chọn C.