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: 72

Answer

$-1$

Work Step by Step

Here, $\lim\limits_{x \to -\infty} f(x)=\lim\limits_{x \to -\infty} \dfrac{2^x+4^x}{5^x-2^x}$ This implies that $\lim\limits_{x \to -\infty} \dfrac{2^x+4^x}{5^x-2^x}(\dfrac{1/2^x}{1/2^x})=\lim\limits_{x \to -\infty} \dfrac{1+(4/2)^x}{(5/2)^x-1}$ Thus, $\dfrac{1+0}{0-1}=-1$
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