Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.3 - The Precise Definition of a Limit - Exercises 2.3 - Page 66: 15


$(3.99,4.01)$ $\delta=0.01$

Work Step by Step

We are given $f(x)=x+1$, $L=5$, $c=4$, and $\epsilon=0.01$. To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$: $$\left|x+1-5 \right| \lt 0.01\\ -0.01 \lt x+1-5 \lt 0.01\\ -0.01 \lt x-4 \lt 0.01\\ 3.99 \lt x \lt 4.01.$$ Thus our interval is $(3.99,4.01)$. Now, to find our $\delta$, we note that $$3.99 \lt x \lt 4.01 \implies -0.01 \lt x-4 \lt 0.01.$$ Hence for $\delta=0.01$, $$0 \lt |x-4| \lt \delta \implies |f(x)-5| \lt \epsilon.$$
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