Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.8 Continuity - 1.8 Exercises - Page 92: 21

Answer

Removable discontinuity at $x=0$

Work Step by Step

$f(x) = \begin{cases} \cos(x)\quad\text{for x<0}\\ 0\quad\quad\quad\text{for x=0}\\ 1-x^2\quad\text{for x>0} \end{cases}$ Let's start by evaluating the left and right hand limits. $\lim\limits_{x \to 0^-}\cos(x) = \cos(0)= 1$ $\lim\limits_{x \to 0^+} = 1-x^2 = 1-(0)^2 = 1$ Since $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+}f(x)$ we can say that $\lim\limits_{x \to 0}f(x)=1$. But $f(0) = 0$. Since $\textbf{the limit exists}$, but $\bf\lim\limits_{x \to 0}f(x)\ne f(0)$ we have a $\textbf{removable or point discontinuity at x=0}$ $\textbf{See graph}$
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