Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.8 Continuity - 1.8 Exercises - Page 94: 71

Answer

$f$ is continuous on $(-\infty,\infty)$.

Work Step by Step

\[f(x)=\left\{\begin{array}{cc}x^4\sin\left(\frac{1}{x}\right)\;\;\;\; x\neq 0\\ 0\;\;\;\;\;\;\;\;\;\; x=0\end{array}\right.\] Clearly $f$ is continuous if $x\neq 0$ so we will check the continuity at $x=0$. Let, \[L=\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0^{-}}x^4\sin\left(\frac{1}{x}\right)\] \[-x^4\leq x^4\sin\left(\frac{1}{x}\right)\leq x^4\] Also $\displaystyle\lim_{x\rightarrow 0}(-x^4)=0=\lim_{x\rightarrow 0}x^4$ By sandwich theorem, \[L=0\] \[\Rightarrow \lim_{x\rightarrow 0}f(x)=f(0)=0\] $\Rightarrow f(x)$ is continuous at $x=0$ Therefore $f$ is continuous on $(-\infty,\infty)$. Hence Proved.
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