Lời giải chi tiết:

Đặt \(h\left( x \right) = \dfrac{{f\left( x \right) - 5}}{{x - 1}} \Leftrightarrow f\left( x \right) = \left( {x - 1} \right)h\left( x \right) + 5\).

\( \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = 5\).

Đặt \(k\left( x \right) = \dfrac{{g\left( x \right) - 1}}{{x - 1}} \Rightarrow g\left( x \right) = \left( {x - 1} \right)k\left( x \right) + 1\).

\( \Rightarrow \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 1\).

Ta có:

\(\begin{array}{l}L = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {f\left( x \right).g\left( x \right) + 4}  - 3}}{{x - 1}}\\L = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right).g\left( x \right) + 4 - 9}}{{\left( {x - 1} \right)\left[ {\sqrt {f\left( x \right).g\left( x \right) + 4}  + 3} \right]}}\\L = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right).g\left( x \right) - 5}}{{\left( {x - 1} \right)\left[ {\sqrt {f\left( x \right).g\left( x \right) + 4}  + 3} \right]}}\\L = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right)\left[ {g\left( x \right) - 1} \right] + f\left( x \right) - 5}}{{\left( {x - 1} \right)\left[ {\sqrt {f\left( x \right).g\left( x \right) + 4}  + 3} \right]}}\\L = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right)\left[ {g\left( x \right) - 1} \right] + \left[ {f\left( x \right) - 5} \right]}}{{\left( {x - 1} \right)\left[ {\sqrt {f\left( x \right).g\left( x \right) + 4}  + 3} \right]}}\\L = \mathop {\lim }\limits_{x \to 1} \dfrac{{g\left( x \right) - 1}}{{x - 1}}.\dfrac{{f\left( x \right)}}{{\sqrt {f\left( x \right).g\left( x \right) + 4}  + 3}} + \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 5}}{{x - 1}}.\dfrac{1}{{\sqrt {f\left( x \right).g\left( x \right) + 4}  + 3}}\\L = 3.\dfrac{5}{{\sqrt {5.1 + 4}  + 3}} + 2.\dfrac{1}{{\sqrt {5.1 + 4}  + 3}}\\L = \dfrac{{15}}{6} + \dfrac{2}{6} = \dfrac{{17}}{6}\end{array}\)

Chọn D.