Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - Review Exercises - Page 331: 57

Answer

$$\frac{1}{{{e^3}}}$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{3}{x}} \right)^x} \cr & {\text{Evaluating}} \cr & \mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{3}{x}} \right)^x} = {1^\infty } \cr & {\text{This limit has the form }}{1^\infty }{\text{ }}\left( {{\text{Using theorem 2}}{\text{.12}}} \right) \cr & {\left( {1 - \frac{3}{x}} \right)^x} = {e^{x\ln \left( {1 - \frac{3}{x}} \right)}},{\text{ then}} \cr & \mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{3}{x}} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {e^{x\ln \left( {1 - \frac{3}{x}} \right)}} = {e^{\mathop {\lim }\limits_{x \to \infty } x\ln \left( {1 - \frac{3}{x}} \right)}} \cr & {\text{The first step is to evaluate }} \cr & L = \mathop {\lim }\limits_{x \to \infty } x\ln \left( {1 - \frac{3}{x}} \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left( {1 - \frac{3}{x}} \right)}}{{1/x}} = \frac{0}{0} \cr & {\text{Using the L'Hopital's rule}} \cr & L = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left( {1 - \frac{3}{x}} \right)}}{{1/x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{\frac{3}{{{x^2}}}}}{{1 - \frac{3}{x}}}}}{{ - \frac{1}{{{x^2}}}}} = - \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{\frac{3}{{{x^2}}}}}{{\frac{{x - 3}}{x}}}}}{{\frac{1}{{{x^2}}}}} = - \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{3x}}{{{x^2}\left( {x - 3} \right)}}}}{{\frac{1}{{{x^2}}}}} \cr & = - \mathop {\lim }\limits_{x \to \infty } \frac{{3{x^3}}}{{{x^2}\left( {x - 3} \right)}} = - \mathop {\lim }\limits_{x \to \infty } \frac{{3x}}{{x - 3}} \cr & - \mathop {\lim }\limits_{x \to \infty } \frac{{3x}}{{x - 3}} = - \mathop {\lim }\limits_{x \to \infty } \frac{3}{{1 - 3/x}} = - 3 \cr & {\text{Therefore,}} \cr & \mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{3}{x}} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {e^{x\ln \left( {1 - \frac{3}{x}} \right)}} = {e^{ - 3}} \cr & \mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{3}{x}} \right)^x} = \frac{1}{{{e^3}}} \cr} $$
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