## Precalculus (6th Edition) Blitzer

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.
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.