Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.5 Derivatives of Trigonometric Functions - 3.5 Exercises - Page 170: 73

Answer

0

Work Step by Step

We have to determine the limit: $L=\lim\limits_{x \to 0}\dfrac{\cos x-1}{x}$ Multiply both numerator and denominator by $\cos x+1$: $L=\lim\limits_{x \to 0}\dfrac{\cos x-1}{x}\cdot\dfrac{\cos x+1}{\cos x+1}=\lim\limits_{x \to 0}\dfrac{\cos^2 x-1}{x(\cos x+1)}$ Use the identity: $\sin^2 x+\cos^2 x=1$ $L=\lim\limits_{x \to 0}\dfrac{-\sin^2 x}{x(\cos x+1)}$ $=-\lim\limits_{x \to 0}\dfrac{\sin^2 x}{x^2}\cdot\dfrac{x}{\cos x+1}=-\lim\limits_{x \to 0}\left(\dfrac{\sin x}{x}\right)^2\cdot \dfrac{x}{\cos x+1}$ Use $\lim\limits_{x \to 0} \dfrac{\sin x}{x}=1$ $L=-\lim\limits_{x \to 0} \dfrac{x}{\cos x+1}=-\dfrac{0}{1+1}=0$
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.