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: 23

Answer

$$\lim_{t\to1}\frac{t^8-1}{t^5-1}=\frac{8}{5}$$

Work Step by Step

$$A=\lim_{t\to1}\frac{t^8-1}{t^5-1}$$ We see that $\lim_{t\to1}(t^8-1)=1^8-1=0$ and $\lim_{t\to1}(t^5-1)=1^5-1=0$, so this is an indeterminate form of $\frac{0}{0}$, which means we can apply L'Hospital's Rule: $$A=\lim_{t\to1}\frac{(t^8-1)'}{(t^5-1)'}$$ $$A=\lim_{t\to1}\frac{8t^7}{5t^4}$$ $$A=\frac{8\times1^7}{5\times1^4}$$ $$A=\frac{8}{5}$$
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