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: 31


1) $L=\lim_{x\to3}f(x)=-3$ 2) $\delta=0.01$

Work Step by Step

$$f(x)=3-2x\hspace{1cm}c=3\hspace{1cm}\epsilon=0.02$$ 1) Find $L=\lim_{x\to3}f(x)$ $$\lim_{x\to3}f(x)=\lim_{x\to3}(3-2x)=3-2\times3=3-6=-3$$ Therefore, $L=\lim_{x\to3}f(x)=-3$ 2) Find $\delta\gt0$ such that for all $x$, $0\lt |x-3|\lt\delta$ then $|f(x)-(-3)|\lt0.02$ - First, solve the inequality to figure out the open interval on which the inequality holds. $$|f(x)-(-3)|\lt0.02$$ $$|3-2x+3|\lt0.02$$ $$|6-2x|\lt0.02$$ $$-0.02\lt6-2x\lt0.02$$ $$-6.02\lt-2x\lt5.98$$ $$3.01\gt x\gt2.99$$ The open interval on which the inequality holds is $(2.99,3.01)$. - Find a value of $\delta\gt0$ The distance from $3$ to both endpoints is the same, which is $0.01$. So if we take $\delta=0.01$, then $0\lt|x-3|\lt0.01$, meaning all $x$ now are placed between $2.99$ and $3.01$, hence $|f(x)-(-3)|\lt0.02$ In other words, $$0\lt|x-3|\lt0.01\Rightarrow|f(x)-(-3)|\lt0.02$$
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