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 76: 30


1) The open interval is $(1-\frac{0.05}{m},1+\frac{0.05}{m})$ 2) $\delta=\frac{0.05}{m}$

Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-1|\lt\delta\Rightarrow|(mx+b)-(m+b)|\lt 0.05\hspace{1cm}m\gt0$$ 1) Find the interval around $1$ on which $|(mx+b)-(m+b)|\lt 0.05$ holds. Solve the inequality: $$|(mx+b)-(m+b)|\lt 0.05$$ $$-0.05\lt mx+b-m-b\lt0.05$$ $$-0.05\lt mx-m\lt0.05$$ $$-\frac{0.05}{m}\lt x-1\lt\frac{0.05}{m}\hspace{1cm}(m\gt0)$$ $$1-\frac{0.05}{m}\lt x\lt1+\frac{0.05}{m}$$ The open interval around $1$ is $(1-\frac{0.05}{m},1+\frac{0.05}{m})$. 2) Give a value for $\delta$ Both endpoints are equally distant from $1$, the distance of which is $\frac{0.05}{m}$ So if we take $\delta=\frac{0.05}{m}$ (since $m\gt0$, $\delta=\frac{0.05}{m}\gt0$) or any smaller positive number, then $0\lt|x-1|\lt\frac{0.05}{m}$, meaning all $x$ would be placed in the interval $(1-\frac{0.05}{m},1+\frac{0.05}{m})$ so that $|(mx+b)-(m+b)|\lt 0.05$. In other words, $$0\lt |x-1|\lt\frac{0.05}{m}\Rightarrow|(mx+b)-(m+b)|\lt 0.05$$
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