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 2.4 - One-Sided Limits - Exercises 2.4 - Page 74: 6


a. Yes. $0$ b. No. c. No.

Work Step by Step

Use the figure in the Exercise, given the function: $g(x)=\sqrt x sin\frac{1}{x}$ a. Yes. $\lim_{x\to0^+}g(x)=\lim_{x\to0^+}\sqrt x sin\frac{1}{x}=0$ because $\lim_{x\to0^+}\sqrt x =0$ and $sin\frac{1}{x}$ oscillates within $[-1,1]$ b. No. $\lim_{x\to0^-}g(x)$ does not exist because the function is undefined below zero. c. No. because $\lim_{x\to0^-}g(x)$ does not exist.
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.