Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 28

Answer

$-\ln 2$

Work Step by Step

Let $\lim\limits_{\theta \to 0}f(x)=\lim\limits_{\theta \to 0} \dfrac{(\dfrac{1}{2})^{ \theta}-1}{\theta}$ But $f(0)=\dfrac{0}{0}$ This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$ This implies that $\lim\limits_{\theta \to 0} \dfrac{(-\dfrac{1}{2})^{ \theta} (\ln 2)}{1}=\dfrac{(-\dfrac{1}{2})^{ 0} (\ln 2)}{1}$ Thus, $(-1) (\ln 2)=-\ln 2$
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