Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 500: 24



Work Step by Step

Given: $\lim\limits_{t\to0}\dfrac{8^t-5^t}{t}$ Here, $\lim\limits_{t\to0}(8^t-5^t)=8^0-5^0=1-1=0$ and $\lim\limits_{t\to0}(t)=0$ This shows an indeterminate form of type $\dfrac{\infty}{-\infty}$, and so we will apply L'Hospital Rule. $\lim\limits_{t\to0}\dfrac{(8^t-5^t)'}{t'}=\lim\limits_{t\to0}\dfrac{8^t\ln8-5^t\ln5}{1}$ or, $=\lim\limits_{t\to0}(8^t\ln8-5^t\ln5)$ or, $=8^0\ln8-5^0\ln5$ or,$=(1)\ln8-(1)\ln5$ or, $=\ln8-\ln5$ or, $=\ln\dfrac{8}{5}$
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