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: 17


The solution is $$\lim_{x\to e}\frac{\ln x-1}{x-e}=\frac{1}{e}.$$

Work Step by Step

We will apply the L'Hopital's rule to solve this limit. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to e}\frac{\ln x-1}{x-e}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to e}\frac{(\ln x-1)'}{(x-e)'}=\lim_{x\to e}\frac{\frac{1}{x}}{1}=\lim_{x\to e}\frac{1}{x}=\frac{1}{e}.$$
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