Answer
$\lim\limits_{x\to0}\dfrac{\sin{x}}{5x}=\dfrac{1}{5}.$
Work Step by Step
Using Theorem $1.9$:
$\lim\limits_{x\to0}\dfrac{\sin{x}}{5x}=\frac{1}{5}\lim\limits_{x\to0}\dfrac{\sin{x}}{x}=\frac{1}{5}(1)=\dfrac{1}{5}.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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 68: 63
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
