## 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 $\underset{x\to 4}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=-5$ " makes sense.
According to the limit property, $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$. So, here $\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)=0$ and $\underset{x\to 4}{\mathop{\lim }}\,g\left( x \right)=-5$. Thus, \begin{align} & \underset{x\to 4}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to 4}{\mathop{\lim }}\,g\left( x \right) \\ & =0+\left( -5 \right) \\ & =-5 \end{align}. Thus, the given statement makes sense.