Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.4 Exercises: 7

Answer

$\lim\limits_{x\to8^+}\dfrac{1}{x+8}=\dfrac{1}{16}.$

Work Step by Step

Using direct substitution: $\lim\limits_{x\to8^+}\dfrac{1}{x+8}=\dfrac{1}{8+8}=\dfrac{1}{16}.$
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.