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 - Page 312: 70

Answer

On the graph, we can see that $~~\frac{5^x-4^x}{3^x-2^x} \approx 0.55~~$ as $~~x \to 0$ $\lim\limits_{x \to 0}\frac{5^x-4^x}{3^x-2^x} = \frac{ln~5-ln~4}{ln~3-ln~2} \approx 0.55034$

Work Step by Step

On the graph, we can see that $~~\frac{5^x-4^x}{3^x-2^x} \approx 0.55~~$ as $~~x \to 0$ $\lim\limits_{x \to 0}\frac{5^x-4^x}{3^x-2^x} = \lim\limits_{x \to 0}\frac{1-1}{1-1} = \frac{0}{0}$ We can apply L'Hospital's Rule. $\lim\limits_{x \to 0}\frac{ (ln~5)~5^x-(ln~4)~4^x }{ (ln~3)~3^x-(ln~2)~2^x} = \frac{ (ln~5)~(1)-(ln~4)~(1) }{ (ln~3)~(1)-(ln~2)~(1)} = \frac{1.609438-1.386294}{1.098612-0.693147} = 0.55034$ Therefore: $\lim\limits_{x \to 0}\frac{5^x-4^x}{3^x-2^x} = \frac{ln~5-ln~4}{ln~3-ln~2} \approx 0.55034$
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