Biết rằng \(\mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 5}}{{x - 1}} = 2\) và \(\mathop {\lim }\limits_{x \to 1} \dfrac{{g\left( x \right) - 1}}{{x - 1}} = 3\). Tính \(\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {f\left( x \right).g\left( x \right) + 4} - 3}}{{x - 1}}\).
A.\(7\)
B.\(17\)
C.\(\dfrac{{23}}{7}\)
D.\(\dfrac{{17}}{6}\)