University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 248: 30


$\dfrac{\ln 3}{\ln 2}$

Work Step by Step

Consider: $\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\dfrac{3^x-1}{2^x-1}$ WenNeed to check that the limit has an indeterminate form. Thus, $f(0)=\dfrac{3^{0}-1}{2^{0}-1}=\dfrac{0}{0}$ The limit shows an indeterminate form. Thus, apply L-Hospital's rule: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ Then $\lim\limits_{x \to 0}\dfrac{3^x \ln 3}{2^x \ln 2}=\dfrac{3^{0} \ln 3}{2^{0} \ln 2}=\dfrac{\ln 3}{\ln 2}$
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