University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.4 - One-Sided Limits - Exercises - Page 84: 6


(a) $\lim_{x\to0^+}f(x)=0$ (b) $\lim_{x\to0^-}f(x)$ does not exist. (c) $\lim_{x\to0}f(x)$ does not exist

Work Step by Step

(a) $\lim_{x\to0^+}f(x)$ Looking at the graph, we cannot see clearly whether as $x\to0$ from the right, $f(x)$ approaches a single value or not. However, we know that $$-1\le\sin\frac{1}{x}\le1$$ $$-\sqrt x\le\sqrt x\sin\frac{1}{x}\le\sqrt x$$ And $\lim_{x\to0^+}(-\sqrt x)=\lim_{x\to0^+}\sqrt x=\sqrt0=0$ Therefore, according to Sandwich Theorem, $\lim_{x\to0^+}f(x)=0$ (b) $\lim_{x\to0^-}f(x)$ does not exist. $f(x)$ is not defined as $x\lt0$, meaning there are not any values of $f(x)$ as $x$ approaches $0$ from the left. (c) Since $\lim_{x\to0^+}f(x)$ exists but $\lim_{x\to0^-}f(x)$ does not exist, $\lim_{x\to0}f(x)$ cannot exist.
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