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: 42

Answer

$cos(a)$

Work Step by Step

$\lim\limits_{x \to a^{+}} cos(x) \times \lim\limits_{x \to a^{+}} \frac{ln(x-a)}{ln(e^x - e^a)} =$ We get $\infty/\infty$, so we apply L'Hospital's Rule: $\lim\limits_{x \to a^{+}} cos(x) \times \lim\limits_{x \to a^{+}} \frac{\frac{1}{x-a}}{\frac{e^x}{e^x - e^a}} =$ $\lim\limits_{x \to a^{+}} cos(x) \times \lim\limits_{x \to a^{+}} \frac{1}{x-a} \times \frac{e^x - e^a}{e^x} =$ $\lim\limits_{x \to a^{+}} cos(x) \times \lim\limits_{x \to a^{+}} \frac{e^x - e^a}{x-a} \times \frac{1}{e^x} =$ $\lim\limits_{x \to a^{+}} cos(x) \times \lim\limits_{x \to a^{+}} \frac{e^x - e^a}{x-a} \times \lim\limits_{x \to a^{+}} \frac{1}{e^x} =$ $cos(a) \times \lim\limits_{x \to a^{+}} \frac{e^x - e^a}{x-a} \times \frac{1}{e^a} =$ We get $0/0$, so we apply L'Hospital's Rule: $cos(a) \times \lim\limits_{x \to a^{+}} \frac{e^x}{1} \times \frac{1}{e^a} =$ $cos(a) \times \lim\limits_{x \to a^{+}} e^a \times \frac{1}{e^a} = cos(a)$
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