Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises - Page 307: 40


The solution is $$\lim_{x\to\infty}\frac{e^{3x}}{3e^{3x}+5}=\frac{1}{3}$$

Work Step by Step

To solve this limit follow the steps below $$\lim_{x\to\infty}\frac{e^{3x}}{3e^{3x}+5}=\lim_{x\to\infty}\frac{e^{3x}}{e^{3x}\left(3+\frac{5}{e^{3x}}\right)}=\lim_{x\to\infty}\frac{1}{3+5e^{-3x}}=\left[\frac{1}{3+5e^{-3\cdot\infty}}\right]=\left[\frac{1}{3+0}\right]=\frac{1}{3}.$$
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