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 65: 10



Work Step by Step

We see from the graph that in order for $f(x)$ to be within $\epsilon=0.2$ of $L=4$, we must have $$2.61 \lt x \lt 3.41.$$ Subtracting $c=3$ from all three sides gives $$-0.39 \lt x-3 \lt 0.41.$$ Note that $$-0.39 \lt x-3 \lt 0.39 \implies -0.39 \lt x-3 \lt 0.41.$$ Hence for $\delta = 0.39$, $$0 \lt |x-3| \lt \delta \implies 0 \lt |f(x)-4| \lt \epsilon.$$
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