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

Answer

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.
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