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: 14



Work Step by Step

We see from the graph that in order for $f(x)$ to be within $\epsilon=0.01$ of $L=2$, we must have $$\dfrac{1}{2.01} \lt x \lt \dfrac{1}{1.99}.$$ Subtracting $c=\dfrac{1}{2}$ from all three sides gives $$-\dfrac{0.01}{4.02} \lt x-\dfrac{1}{2} \lt \dfrac{0.01}{3.98}.$$ $$-\dfrac{1}{402} \lt x-\dfrac{1}{2} \lt \dfrac{1}{398}.$$ Note that $$-\dfrac{1}{402} \lt x-\dfrac{1}{2} \lt \dfrac{1}{402} \implies -\dfrac{1}{402} \lt x-\dfrac{1}{2} \lt \dfrac{1}{398}.$$ Hence for $\delta = \dfrac{1}{402}$, $$0 \lt \left|x-\dfrac{1}{2}\right| \lt \delta \implies 0 \lt |f(x)-2| \lt \epsilon.$$
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