Nếu \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 5}}{{x - 1}} = 2\) và \(\mathop {\lim }\limits_{x \to 1} \frac{{g\left( x \right) - 1}}{{x - 1}} = 3\) thì \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f\left( x \right) \cdot g\left( x \right) + 4} - 3}}{{x - 1}}\) bằng:

Đặt \(h\left( x \right) = \frac{{f\left( x \right) - 5}}{{x - 1}} \Rightarrow \left\{ \begin{array}{l}f\left( x \right) = \left( {x - 1} \right)h\left( x \right) + 5\\\mathop {\lim }\limits_{x \to 1} \,\,h\left( x \right) = 2\end{array} \right.;\)\(p\left( x \right) = \frac{{g\left( x \right) - 1}}{{x - 1}} \Rightarrow \left\{ \begin{array}{l}g\left( x \right) = \left( {x - 1} \right)p\left( x \right) + 1\\\mathop {\lim }\limits_{x \to 1} \,\,p\left( x \right) = 3\end{array} \right..\)

Ta có \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f\left( x \right) \cdot g\left( x \right) + 4}  - 3}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) \cdot g\left( x \right) - 5}}{{\left( {x - 1} \right)\left( {\sqrt {f\left( x \right) \cdot g\left( x \right) + 4}  + 3} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 1} \frac{{{{\left( {x - 1} \right)}^2}h\left( x \right)p\left( x \right) + \left( {x - 1} \right)\left[ {h\left( x \right) + 5p\left( x \right)} \right]}}{{\left( {x - 1} \right)\left[ {\sqrt {{{\left( {x - 1} \right)}^2}h\left( x \right)p\left( x \right) + \left( {x - 1} \right)\left[ {h\left( x \right) + 5p\left( x \right)} \right] + 9}  + 3} \right]}}\)

\( = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)h\left( x \right)p\left( x \right) + \left[ {h\left( x \right) + 5p\left( x \right)} \right]}}{{\sqrt {{{\left( {x - 1} \right)}^2}h\left( x \right)p\left( x \right) + \left( {x - 1} \right)\left[ {h\left( x \right) + 5p\left( x \right)} \right] + 9}  + 3}} = \frac{{17}}{6}\). Chọn A.