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 85: 35


\begin{aligned} \lim _{t \rightarrow 0} \cos \left(\frac{\pi}{\sqrt{19-3 \sec 2 t}}\right) =\frac{\sqrt{2}}{2} \end{aligned} The function is continuous at $ t=0$.

Work Step by Step

Given $$ \lim _{t \rightarrow 0} \cos \left(\frac{\pi}{\sqrt{19-3 \sec 2 t}}\right)$$ \begin{aligned} \lim _{t \rightarrow 0} \cos \left(\frac{\pi}{\sqrt{19-3 \sec 2 t}}\right) &=\cos \left(\frac{\pi}{\sqrt{19-3 \sec 0}}\right)\\ &=\cos \left(\frac{\pi}{\sqrt{19-3}}\right)\\ &=\cos \left(\frac{\pi}{4}\right)\\ &=\frac{\sqrt{2}}{2} \end{aligned} Since $$ f(t)= \cos \left(\frac{\pi}{\sqrt{19-3 \sec 2 t}}\right)$$ \begin{aligned}f(0)&=\cos \left(\frac{\pi}{\sqrt{19-3 \sec 0}}\right) \\ &=\cos \left(\frac{\pi}{\sqrt{19-3}}\right)\\ &=\cos \left(\frac{\pi}{4}\right)\\ &=\frac{\sqrt{2}}{2} \end{aligned} From (a), (b) since $\lim \limits_{t \rightarrow 0} f(t)=f(0)=\frac{\sqrt{2}}{2},$ the function is continuous at $ t=0$.
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