Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.7 Limits at Infinity - Exercises - Page 83: 34


(a) $286~K$ (b) $25.94~years$

Work Step by Step

We are given the function $$ 283+3\left(1-10^{-0.03 t}\right)$$ (a) We have the limit: \begin{aligned} \lim _{t \rightarrow \infty} 283+3\left(1-10^{-0.03 t}\right) &=283+3\left(1-\lim _{t \rightarrow \infty}\left(10^{-0.03 t}\right)\right) \\ &=283+3(1-0) \\ &=286~K \end{aligned} (b) We find $t$, when $T=286-0.5$ \begin{aligned} 283+3\left(1-10^{-0.03 t}\right)&\geq 285.5\\ 1-10^{-0.03 t} &\geq \frac{2.5}{3}\\ -10^{-0.03 t} &\geq-\frac{1}{6}\\ 10^{-0.03 t}& \leq \frac{1}{6}\\ -0.03 t &\leq \log_{10} \frac{1}{6} \end{aligned} Then $$ t \approx 25.94~yrs$$
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