Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises - Page 103: 35

Answer

Squeeze Theorem $lim _{x\to 0}(x^2cos20\pi x)=0$

Work Step by Step

Squeeze Theorem If $h\left(x\right)\le f\left(x\right)\le g\left(x\right)$ If $lim _{x\to a}h\left(x\right)=L$ and $lim _{x\to a}g\left(x\right)=L$ then: $lim _{x\to a}f\left(x\right)=L$ We know $-1\le cos20\pi x\le 1$ Multiply by $x^{2}$: $-x^2\le x^2 cos20\pi x\le x^2$ Since $lim _{x\to 0}\left(-x^2\right)=0$ $lim _{x\to 0}\left(x^2\right)=0$ By Squeeze Theorem $lim _{x\to 0}(x^2cos20\pi x)=0$ See graph.
Small 1512390573
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.