Phương pháp giải:

- Tính \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).

- Phân tích  giới hạn \(\mathop {\lim }\limits_{x \to 2} \left[ {\dfrac{{f\left( x \right) - 15}}{{x - 2}}.\dfrac{1}{{\left( {x + 2} \right)\left( {\sqrt {2f\left( x \right) + 6}  + 3} \right)}}} \right]\), sau đó tính giới hạn.

Lời giải chi tiết:

Đặt \(g\left( x \right) = \dfrac{{f\left( x \right) - 15}}{{x - 2}}\) \( \Rightarrow f\left( x \right) = \left( {x - 2} \right)g\left( x \right) + 15\).

\( \Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left[ {\left( {x - 2} \right)g\left( x \right) + 15} \right] = 15\).

Ta có:

\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 15}}{{\left( {{x^2} - 4} \right)\left( {\sqrt {2f\left( x \right) + 6}  + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 15}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {\sqrt {2f\left( x \right) + 6}  + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \left[ {\dfrac{{f\left( x \right) - 15}}{{x - 2}}.\dfrac{1}{{\left( {x + 2} \right)\left( {\sqrt {2f\left( x \right) + 6}  + 3} \right)}}} \right]\\ = 3.\dfrac{1}{{4.\left( {\sqrt {2.15 + 6}  + 3} \right)}} = 3.\dfrac{1}{{4.9}} = \dfrac{1}{{12}}\end{array}\)

Chọn C.