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: 19

Answer

The limit does not exist (jump discontinuity)

Work Step by Step

$f(x)= \begin{cases} 1-x^2\quad \text{for }x\lt1\\ \dfrac{1}{x} \qquad\text{ for }x \geq1 \\ \end{cases} $ We can test what kind of discontinuity we might have by first looking at the left and right hand limits. $\lim\limits_{x \to 1^-} 1-x^2 = 1-(1^2) =0$ $\lim\limits_{x \to 1^+} \dfrac{1}{x} = \dfrac{1}{1} =1$ Because $\lim\limits_{x \to 1^-}f(x) \ne\lim\limits_{x \to 1^+}f(x)\quad\textbf{the limit does not exist.}$ So we can see from this that the function jumps from $0$ to $1$ at $x=1$. So we have a $\textbf{jump discontinuity at x=1}$ We can graph this manually on paper. For the first piece, $f(x)=1-x^2$, try plugging in values $0,-1$, and $-2$. For the second piece, $f(x)= \dfrac{1}{x}$, try plugging in values $2, 3$, and $4$. $\textbf{See graph}$
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