Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.3 Techniques for Computing Limits - 2.3 Exercises - Page 78: 65



Work Step by Step

$\lim _{h\rightarrow 0}\dfrac {\left( 5+h\right) ^{2}-25}{h}=\dfrac {25+10h+h^{2}-25}{h}=\dfrac {10h+h^{2}}{h}=\dfrac {h\left( 10+h\right) }{h}=10+h=10+0=10$
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.