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 - 1.3 Exercises - Page 67: 48

Answer

$\lim\limits_{x\to 0}\dfrac{2x}{x^2+4x}=\dfrac{1}{2}.$

Work Step by Step

$\lim\limits_{x\to 0}\dfrac{2x}{x^2+4x}=\lim\limits_{x\to 0}\dfrac{x(2)}{x(x+4)}=\lim\limits_{x\to 0}\dfrac{2}{x+4}=\dfrac{2}{0+4}=\dfrac{1}{2}.$
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.