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 - Exercise Set - Page 1161: 39

Answer

There is no number for which the function $ f\left( x \right)=\left\{ \begin{align} & \frac{\sin 2x}{x}\text{if }x\ne 0 \\ & 2\text{ if }x=0 \end{align} \right.$ is discontinuous.

Work Step by Step

Consider the function $ f\left( x \right)=\left\{ \begin{align} & \frac{\sin 2x}{x}\text{if }x\ne 0 \\ & 2\text{ if }x=0 \end{align} \right.$, Check the discontinuity of the function at $ x=0$ Find the value of $ f\left( x \right)$ at $ x=0$, From the definition of the function, $ f\left( 0 \right)=2$ The function is defined at the point $ x=0$. Now find the value of $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$, Take some points near to the left of $0$ Suppose the points are $-0.03,-0.02\text{ and }-0.01$ Take some points near to the right of $0$ Suppose the points are $0.01,0.02\text{ and }0.03$ Now evaluate the values of the function at the above chosen points. As x nears $0$ from the left or right the value of the function nears $2$ Thus $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=2$ From the above steps, $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=2=f\left( 0 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function $ f\left( x \right)=\left\{ \begin{align} & \frac{\sin 2x}{x}\text{if }x\ne 0 \\ & 2\text{ if }x=0 \end{align} \right.$ is continuous at $0$. Thus, there is no number for which the function $ f\left( x \right)=\left\{ \begin{align} & \frac{\sin 2x}{x}\text{if }x\ne 0 \\ & 2\text{ if }x=0 \end{align} \right.$ is discontinuous.
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