Answer
$=\dfrac {-2}{3}$
Work Step by Step
$\lim _{x\rightarrow -4}\dfrac {f\left( x\right) +1}{3g\left( x\right) -9}=\dfrac {\lim _{x\rightarrow -4}f\left( x\right) +1}{3\lim _{x\rightarrow -4}g\left( x\right) -9}=\dfrac {3+1}{3\times 1-9}=\dfrac {4}{-6}=\dfrac {-2}{3}$
