Answer
$$ - \frac{{137}}{8}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 4} \frac{{5g\left( x \right) + 2}}{{1 - f\left( x \right)}} \cr
& {\text{use the rule 4 for limits }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}{\text{ }}\left( {{\text{see page 128}}} \right) \cr
& = \frac{{\mathop {\lim }\limits_{x \to 4} \left[ {5g\left( x \right) + 2} \right]}}{{\mathop {\lim }\limits_{x \to 4} \left[ {1 - f\left( x \right)} \right]}} \cr
& {\text{use rule rule 1 and rule 2 }} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 4} 5g\left( x \right) + \mathop {\lim }\limits_{x \to 4} 2}}{{\mathop {\lim }\limits_{x \to 4} 1 - \mathop {\lim }\limits_{x \to 4} f\left( x \right)}} \cr
& = \frac{{5\mathop {\lim }\limits_{x \to 4} g\left( x \right) + \mathop {\lim }\limits_{x \to 4} 2}}{{\mathop {\lim }\limits_{x \to 4} 1 - \mathop {\lim }\limits_{x \to 4} f\left( x \right)}} \cr
& {\text{substitute the known limits }}\mathop {\lim }\limits_{x \to 4} g\left( x \right) = 27{\text{ and }}\mathop {\lim }\limits_{x \to 4} f\left( x \right) = 9 \cr
& {\text{ and use }}\mathop {\lim }\limits_{x \to a} k = k \cr
& = \frac{{5\left( {27} \right) + 2}}{{1 - 9}} \cr
& = \frac{{137 + 2}}{{ - 8}} \cr
& = - \frac{{137}}{8} \cr} $$
