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 2.5 - Continuity - Exercises 2.5 - Page 84: 34


$$\lim\limits _{x \rightarrow 0} \tan \left(\frac{\pi}{4} \cos \left(\sin x^{1 / 3}\right)\right)=\frac{1}{\sqrt 2} $$ The function is continuous at $ x=0$.

Work Step by Step

Given $$\lim\limits _{x \rightarrow 0} \tan \left(\frac{\pi}{4} \cos \left(\sin x^{1 / 3}\right)\right) $$ So, \begin{aligned}a) L&=\lim\limits _{x \rightarrow 0} \tan \left(\frac{\pi}{4} \cos \left(\sin x^{1 / 3}\right)\right)\\ &= \tan \left(\frac{\pi}{4} \cos \left(\sin 0\right)\right)\\ &= \tan \left(\frac{\pi}{4} \cos0\right)\\ &= \tan \left(\frac{\pi}{4} \right)\\ &=\frac{1}{\sqrt 2}\\ \end{aligned} Since $$ f(x)=\tan \left(\frac{\pi}{4} \cos \left(\sin x^{1 / 3}\right)\right)$$ \begin{aligned}b) f(0)&= \tan \left(\frac{\pi}{4} \cos \left(\sin 0\right)\right)\\ &= \tan \left(\frac{\pi}{4} \cos0\right)\\ &= \tan \left(\frac{\pi}{4} \right)\\ &=\frac{1}{\sqrt 2}\\ \end{aligned} From (a), (b) since $\lim \limits_{x \rightarrow 0} f(x)=f(0)=\frac{1}{\sqrt 2},$ the function is continuous at $ x=0$.
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