Precalculus (6th Edition) Blitzer

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

Chapter 11 - Section 11.3 - Limits and Continuity - Concept and Vocabulary Check - Page 1160: 3


The statement β€œFor the piecewise function defined by $ f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-25}{x-5}\text{ if }x\ne 5 \\ & \text{12 if }x=5\text{ } \end{align} \right.$, $\,\underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)$ exists and is equal to $10$, so f is continuous at $5$” is false.

Work Step by Step

Consider the function, $ f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-25}{x-5}\text{ if }x\ne 5 \\ & \text{12 if }x=5\text{ } \end{align} \right.$ For a function to be continuous at a point a, the function must satisfy the following three conditions: (a) f is defined at a. (b) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists. (c) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ From the definition of the function, $ f\left( 5 \right)=12$ The function is defined at $5$. Thus, $\,\underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)$ exists and is equal to $10$. Thus, $\,\underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)=10$ From the above two steps, $\,\underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)\ne f\left( 5 \right)$ Thus, the function do not satisfy the third condition of being continuous. Therefore, the function f is not continuous at $5$. Hence, the statement is false.
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