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: 27

Answer

$\lim\limits_{\Delta x\to0}\dfrac{\sin{(\frac{\pi}{6}+\Delta x)}-\frac{1}{2}}{\Delta x}=\dfrac{\sqrt{3}}{2}.$

Work Step by Step

$\lim\limits_{\Delta x\to0}\dfrac{\sin{(\frac{\pi}{6}+\Delta x)}-\frac{1}{2}}{\Delta x}$ $=\lim\limits_{\Delta x\to0}\dfrac{\sin{\frac{\pi}{6}}\cos{\Delta x}+\cos{\frac{\pi}{6}}\sin{\Delta x}-\frac{1}{2}}{\Delta x}$ $=-\frac{1}{2}(\lim\limits_{\Delta x\to0}\dfrac{1-\cos{\Delta x}}{\Delta x})+\frac{\sqrt{3}}{2}(\lim\limits_{\Delta x\to0}\dfrac{\sin{\Delta x}}{x})$ $=-\frac{1}{2}(0)+\frac{\sqrt{3}}{{2}}(1)=\dfrac{\sqrt{3}}{2}.$
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.