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: 13



Work Step by Step

We see from the graph that in order for $f(x)$ to be within $\epsilon=0.5$ of $L=2$, we must have $$-\dfrac{16}{9} \lt x \lt -\dfrac{16}{25}.$$ Subtracting $c=-1$ from all three sides gives $$-\dfrac{7}{9} \lt x-(-1) \lt \dfrac{9}{25}.$$ Note that $$-\dfrac{9}{25} \lt x-(-1) \lt \dfrac{9}{25} \implies -\dfrac{7}{9} \lt x-(-1) \lt \dfrac{9}{25}.$$ Hence for $\delta = \dfrac{9}{25}$, $$0 \lt |x-(-1)| \lt \delta \implies 0 \lt |f(x)-2| \lt \epsilon.$$
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