Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.5 - Continuity - Exercises 2.5 - Page 85: 37


$0$, discontinuous.

Work Step by Step

Step 1. Simplify the rational function under the square root: $\frac{cos^2x-cos(x)}{x}=\frac{cos(x)(cos(x)-1)}{x}=\frac{cos(x)(-2sin^2(x/2))}{x}=-cos(x)sin(x/2)\frac{sin(x/2)}{x/2}$ Step 2. Since $\lim_{x\to0}\frac{sin(x/2)}{x/2}=1$, $\lim_{x\to0}sin(x/2)=0$, and $\lim_{x\to0}\cos(x)=1$, we have $\lim_{x\to0}\frac{cos^2x-cos(x)}{x}=0$, Step 3. We can reach a result that $\lim_{x\to0}sin \sqrt {|\frac{cos^2x-cos(x)}{x}|}=0$, Step 4. The function is not continuous at $x=0$ as it is not defined at this point (dividing by zero). Please note: we need to put absolute signs under the square root to avoid negative values.
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.