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: 7


The complete statement is “$\underset{x\to a}{\mathop{\lim }}\,\text{ }{{\left[ f\left( x \right) \right]}^{n}}=$${{L}^{n}}$ where $ n\ge 2$ is an integer”.

Work Step by Step

In case of the limit of a power, find the limit of the function and then raise the limit to the power. That is: The limit of a power: If $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L $ and $ n $ is a positive integer, then $\underset{x\to a}{\mathop{\lim }}\,{{\left[ f\left( x \right) \right]}^{n}}={{\left[ \underset{x\to a}{\mathop{\lim }}\,f\left( x \right) \right]}^{n}}={{L}^{n}}$ The limit of the function $ f\left( x \right)$ as $ x $ approaches $ a $ is the function evaluated at $ a $. Because this function is raised to the $ n\text{th}$ power, the limit that we seek is the limit of the function raised to the $ n\text{th}$ power. For example: Let $ f\left( x \right)=x $, $\begin{align} & \underset{x\to 7}{\mathop{\lim }}\,{{\left[ f\left( x \right) \right]}^{2}}={{\left[ \underset{x\to 7}{\mathop{\lim }}\,f\left( x \right) \right]}^{2}} \\ & ={{\left[ \underset{x\to 7}{\mathop{\lim }}\,x \right]}^{2}} \\ & ={{\left[ 7 \right]}^{2}} \\ & =49 \end{align}$ Therefore, the complete fill for the blank in the statement “ $\underset{x\to a}{\mathop{\lim }}\,\text{ }{{\left[ f\left( x \right) \right]}^{n}}=$${{L}^{n}}$, where $ n\ge 2$ is an integer”.
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