University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 75: 3


We take $\delta=1/2$ here.

Work Step by Step

$$a=-7/2 \hspace{1cm} b=-1/2\hspace{1cm}c=-3$$ Here we need to find $\delta\gt0$ such that for all $x$, $0\lt |x-(-3)|\lt\delta$, or $0\lt|x+3|\lt\delta$, then $-7/2\lt x\lt -1/2$. The sketch is shown below. The nearer endpoint of $(-7/2,-1/2)$ to $c=-3$ to $-7/2$, and the distance is $1/2$. So if we take $\delta=1/2$ or any smaller positive value, then $0\lt |x+3|\lt1/2$, and $x$ will only range in $(-7/2,-5/2)$, as shown by the red lines in the sketch, thereby $x$ is always placed between $-7/2$ and $-1/2$ as requested. In other words, $$0\lt |x+3|\lt\frac{1}{2}\hspace{2cm}-\frac{7}{2}\lt x\lt -\frac{1}{2}$$
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