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 - Concept and Vocabulary Check - Page 1153: 6


The complete statement is “$\underset{x\to a}{\mathop{\lim }}\,\text{ }\left[ f\left( x \right)\cdot g\left( x \right) \right]=$$ LM $.”

Work Step by Step

In case of the limit of a product, find the limit of each function in the product and then take the product of each of the limits. That is, The limit of a product: If $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L\text{ and }\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=M $, then $\underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right)\cdot g\left( x \right) \right]\underset{x\to a}{\mathop{=\lim }}\,f\left( x \right)\cdot \underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=LM $. The limit of the product of two functions equals the product of their limits. For example: Let $ f\left( x \right)=x $ and $ g\left( x \right)=2$, $\begin{align} & \underset{x\to 7}{\mathop{\lim }}\,\left[ f\left( x \right)\cdot g\left( x \right) \right]=\underset{x\to 7}{\mathop{\lim }}\,f\left( x \right)\cdot \underset{x\to 7}{\mathop{\lim }}\,g\left( x \right) \\ & =\underset{x\to 7}{\mathop{\lim }}\,x\cdot \underset{x\to 7}{\mathop{\lim }}\,2 \\ & =7\cdot 2 \\ & =14 \end{align}$ Therefore, the complete fill for the blank in the statement “$\underset{x\to a}{\mathop{\lim }}\,\text{ }\left[ f\left( x \right)\cdot g\left( x \right) \right]=$$ LM $.”
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