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: 71



Work Step by Step

Here, $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to\infty} \dfrac{2^x-3^x}{3^x+4^x}$ This implies that $\lim\limits_{x \to\infty} \dfrac{2^x-3^x}{3^x+4^x}(\dfrac{1/3^x}{1/3^x})=\lim\limits_{x \to\infty} \dfrac{(2/3)^x-1}{1+(4/3)^x}$ Thus, $\dfrac{0-1}{1+\infty}=0$
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