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: 48

Answer

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

Work Step by Step

To solve this limit we will use L'Hopital's rule. "LR" will denote "Apply L'Hopital's rule". $$\lim_{x\to\infty}\csc\frac{1}{x}(e^{1/x}-1)=\lim_{x\to\infty}\frac{e^{\frac{1}{x}}-1}{\sin\frac{1}{x}}=\left[\frac{e^{\frac{1}{\infty}}-1}{\sin\frac{1}{\infty}}\right]=\left[\frac{e^0-1}{\sin0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\infty}\frac{\left(e^{\frac{1}{x}}-1\right)'}{\left(\sin\frac{1}{x}\right)'}=\lim_{x\to\infty}\frac{e^{\frac{1}{x}}\left(\frac{1}{x}\right)'}{\cos\frac{1}{x}\left(\frac{1}{x}\right)'}=\lim_{x\to\infty}\frac{e^{\frac{1}{x}}}{\cos\frac{1}{x}}=\left[\frac{e^{1/\infty}}{\cos(1/\infty)}\right]=\frac{e^0}{\cos0}=1.$$
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