Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises - Page 213: 101



Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to e} \frac{{\ln x - 1}}{{x - e}} \cr & {\text{rewrite}} \cr & \mathop {\lim }\limits_{x \to e} \frac{{\ln x - 1}}{{x - e}} = \mathop {\lim }\limits_{x \to e} \frac{{\ln x - \ln \left( e \right)}}{{x - e}} \cr & {\text{using the definition of the derivative }}f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} \cr & {\text{comparing }}\mathop {\lim }\limits_{x \to e} \frac{{\ln x - \ln \left( e \right)}}{{x - e}}{\text{ with the definition of the derivative}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to e} \frac{{\ln x - \ln \left( e \right)}}{{x - e}}:f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to e} \frac{{\ln x - \ln \left( e \right)}}{{x - e}}:f'\left( e \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( e \right)}}{{x - e}} \cr & {\text{we can note that }}f\left( x \right) = \ln x{\text{ and }}a = e \cr & {\text{then}} \cr & \,\,\,\,\,\,\,\,f'\left( x \right) = \frac{1}{x} \cr & \,\,\,\,\,\,\,\,f'\left( e \right) = \frac{1}{e} \cr & {\text{and}} \cr & \,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to e} \frac{{\ln x - 1}}{{x - e}} = f'\left( e \right) = \frac{1}{e} \cr} $$
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