Answer
The statement "I am working with functions $ f $ and $ g $ for which $\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)=0$, $\underset{x\to 4}{\mathop{\lim }}\,g\left( x \right)=-5$, and $\underset{x\to 4}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}=0$ " makes sense.
Work Step by Step
According to quotient property of limits,
$\underset{x\to a}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}=\frac{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}{\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)},\text{ }if\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)\ne 0$
So, here $\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)=0$, $\underset{x\to 4}{\mathop{\lim }}\,g\left( x \right)=-5\ne 0$
Thus, $\begin{align}
& \underset{x\to 4}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}=\frac{\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)}{\underset{x\to 4}{\mathop{\lim }}\,g\left( x \right)} \\
& =\frac{0}{-5} \\
& =0
\end{align}$
Thus, the given statement makes sense.
