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 - Review Exercises: 25

Answer

$\lim\limits_{x\to0}\dfrac{1-\cos{x}}{\sin{x}}=0.$

Work Step by Step

Using Theorem $1.9:$ $\lim\limits_{x\to0}\dfrac{1-\cos{x}}{\sin{x}}=\lim\limits_{x\to0}\dfrac{1-\cos{x}}{x}\times\lim\limits_{x\to0}\dfrac{x}{\sin{x}}=0\times1=0.$
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.