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: 28


1) The open interval is $(3-\frac{c}{m},3+\frac{c}{m})$ 2) $\delta=\frac{c}{m}$

Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-3|\lt\delta\Rightarrow|mx-3m|\lt c\hspace{1cm}m\gt0, c\gt0$$ 1) Find the interval around $3$ on which $|mx-3m|\lt c$ holds. Solve the inequality: $$|mx-3m|\lt c$$ $$-c\lt mx-3m\lt c\hspace{1cm}(c\gt0)$$ $$-\frac{c}{m}\lt x-3\lt\frac{c}{m}\hspace{1cm}(m\gt0)$$ $$3-\frac{c}{m}\lt x\lt 3+\frac{c}{m}$$ The open interval around $3$ is $(3-\frac{c}{m},3+\frac{c}{m})$. 2) Give a value for $\delta$ Both endpoints are equally distant from $3$, the distance of which is $\frac{c}{m}$ So if we take $\delta=\frac{c}{m}$ (since $m\gt0$ and $c\gt0$, $\delta=\frac{c}{m}\gt0$) or any smaller positive number, then $0\lt|x-3|\lt\frac{c}{m}$, meaning all $x$ would be placed in the interval $(3-\frac{c}{m},3+\frac{c}{m})$ so that $|mx-3m|\lt c$. In other words, $$0\lt |x-3|\lt\frac{c}{m}\Rightarrow|mx-3m|\lt c$$
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