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: 12

Answer

$\delta=-\dfrac{2-\sqrt{5}}{2} \gt 0$

Work Step by Step

We see from the graph that in order for $f(x)$ to be within $\epsilon=0.25$ of $L=3$, we must have $$-\dfrac{\sqrt{5}}{2} \lt x \lt -\dfrac{\sqrt{3}}{2}.$$ Subtracting $c=-1$ from all three sides gives $$\dfrac{2-\sqrt{5}}{2} \lt x-(-1) \lt \dfrac{2-\sqrt{3}}{2}.$$ Now, $\dfrac{2-\sqrt{5}}{2} \approx -0.118$ and $\dfrac{2-\sqrt{3}}{2} \approx 0.134$. This means $\left|\dfrac{2-\sqrt{5}}{2}\right| \lt \left|\dfrac{2-\sqrt{3}}{2}\right|$. Thus, $$\dfrac{2-\sqrt{5}}{2} \lt x-(-1) \lt -\dfrac{2-\sqrt{5}}{2} \implies \dfrac{2-\sqrt{5}}{2} \lt x-(-1) \lt \dfrac{2-\sqrt{3}}{2}.$$ Hence for $\delta = -\dfrac{2-\sqrt{5}}{2} \gt 0$, $$0 \lt |x-(-1)| \lt \delta \implies 0 \lt |f(x)-3| \lt \epsilon.$$
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