Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises: 24

Answer

$$\lim_{t\to0}\frac{8^t-5^t}{t}=\ln\frac{8}{5}$$

Work Step by Step

$$A=\lim_{t\to0}\frac{8^t-5^t}{t}$$ Since $\lim_{t\to0}(8^t-5^t)=8^0-5^0=1-1=0$ and $\lim_{t\to0}(t)=0$, so we have an indeterminate form of $\frac{0}{0}$. Now we can use L'Hospital's Rule: $$A=\lim_{t\to0}\frac{(8^t-5^t)'}{t'}$$ $$A=\lim_{t\to0}\frac{8^t\ln8-5^t\ln5}{1}$$ $$A=\lim_{t\to0}(8^t\ln8-5^t\ln5)$$ $$A=8^0\ln8-5^0\ln5$$ $$A=1\times\ln8-1\times\ln5$$ $$A=\ln8-\ln5$$ $$A=\ln\frac{8}{5}$$
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