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: 29

Answer

The solution $$\lim_{x\to\infty}\frac{e^{1/x}-1}{1/x}=1.$$

Work Step by Step

We will calculate this limit using L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule": $$\lim_{x\to\infty}\frac{e^{1/x}-1}{1/x}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\infty}\frac{(e^{1/x}-1)'}{(1/x)'}=\lim_{x\to\infty}\frac{e^{1/x}(1/x)'}{(1/x)'}=\lim_{x\to\infty}e^{1/x}=e^{1/\infty}=e^0=1.$$
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