Answer
$$\frac{2}{3}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 4} \frac{{f\left( x \right) + g\left( x \right)}}{{2g\left( x \right)}} \cr
& {\text{use the rule 4 for limits }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}{\text{ }}\left( {{\text{see page 128}}} \right) \cr
& = \frac{{\mathop {\lim }\limits_{x \to 4} \left[ {f\left( x \right) + g\left( x \right)} \right]}}{{\mathop {\lim }\limits_{x \to 4} 2g\left( x \right)}} \cr
& {\text{use rule rule 1 and rule 2 }} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 4} f\left( x \right) + \mathop {\lim }\limits_{x \to 4} g\left( x \right)}}{{\mathop {2\lim }\limits_{x \to 4} g\left( x \right)}} \cr
& {\text{substitute the known limits }}\mathop {\lim }\limits_{x \to 4} g\left( x \right) = 27{\text{ and }}\mathop {\lim }\limits_{x \to 4} f\left( x \right) = 9 \cr
& = \frac{{9 + 27}}{{2\left( {27} \right)}} \cr
& = \frac{{36}}{{54}} \cr
& = \frac{2}{3} \cr} $$
