Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises - Page 126: 71


The function is continuous on $(-\infty, \infty)$.

Work Step by Step

$f(x) = x^4~sin(\frac{1}{x})$, if $x \neq 0$ Clearly this function is continuous on the intervals $(-\infty, 0)$ and $(0, \infty)$ Since $f(0) = 0$, we need to show that $\lim\limits_{x \to 0}x^4~sin(\frac{1}{x}) = 0$ Let $\epsilon \gt 0$ be given. Let $\delta = \epsilon^{1/4}$ Suppose that $\vert x - 0 \vert \lt \delta$ (but $x \neq 0$) Then: $\vert x^4~sin(\frac{1}{x}) \vert = \vert x^4 \vert \cdot \vert sin(\frac{1}{x})\vert \lt \delta^4 \cdot 1 = \epsilon$ Therefore, $\lim\limits_{x \to 0}x^4~sin(\frac{1}{x}) = 0$ Thus, the function is continuous on $(-\infty, \infty)$
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