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: 11



Work Step by Step

We see from the graph that in order for $f(x)$ to be within $\epsilon=1$ of $L=4$, we must have $$\sqrt{3} \lt x \lt \sqrt{5}.$$ Subtracting $c=2$ from all three sides gives $$\sqrt{3}-2 \lt x-2 \lt \sqrt{5}-2.$$ Now, $\sqrt{3}-2 \approx -0.268$ and $\sqrt{5}-2 \approx 0.236$. This means $|\sqrt{5}-2| \lt |\sqrt{3}-2|$. Thus, $$-(\sqrt{5}-2) \lt x-2 \lt \sqrt{5}-2 \implies \sqrt{3}-2 \lt x-2 \lt \sqrt{5}-2.$$ Hence for $\delta = \sqrt{5}-2$, $$0 \lt |x-2| \lt \delta \implies 0 \lt |f(x)-4| \lt \epsilon.$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.