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 410: 72



Work Step by Step

Here, we have $\lim\limits_{x \to -\infty} f(x)=\lim\limits_{x \to -\infty} \dfrac{2^x+4^x}{5^x-2^x}$ Re-arrange as: $\lim\limits_{x \to -\infty} \dfrac{2^x+4^x}{5^x-2^x}(\dfrac{\dfrac{1}{2^x}}{\dfrac{1}{2^x}})=\lim\limits_{x \to -\infty} \dfrac{1+(\dfrac{4}{2})^x}{(\dfrac{5}{2})^x-1}$ and $\dfrac{1+0}{0-1}=-1$
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