Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.2 - Finding Limits Using Properties of Limits - Exercise Set - Page 1155: 75

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 $\underset{x\to 4}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=-5$ " makes sense.

Work Step by Step

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.
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