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 \r

* Hướng dẫn giải

Vì \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 5}}{{x - 1}} = 2 \Rightarrow f\left( 1 \right) = 5\) và \(\mathop {\lim }\limits_{x \to 1} \frac{{g\left( x \right) - 1}}{{x - 1}} = 3 \Rightarrow g\left( 1 \right) = 1\)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f\left( x \right).g\left( x \right) + 4}  - 3}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right).g\left( x \right) - 5}}{{\left( {x - 1} \right)\left( {\sqrt {f\left( x \right).g\left( x \right) + 4}  + 3} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{f\left( x \right).\left[ {g\left( x \right) - 1} \right]}}{{x - 1}} + \frac{{f\left( x \right) - 5}}{{x - 1}}}}{{\left( {\sqrt {f\left( x \right).g\left( x \right) + 4}  + 3} \right)}}\\
 = \frac{{f\left( 1 \right).3 + 2}}{{\sqrt {f\left( 1 \right).g\left( 1 \right) + 4}  + 3}} = \frac{{5.3 + 2}}{{\sqrt {5 + 4}  + 3}} = \frac{{17}}{6}
\end{array}\)