Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.7 Exercises - Page 564: 54

Answer

$$1$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to {4^ + }} {\left[ {3\left( {x - 4} \right)} \right]^{x - 4}} \cr & {\text{Evaluating the limit}} \cr & \mathop {\lim }\limits_{x \to {4^ + }} {\left[ {3\left( {x - 4} \right)} \right]^{x - 4}} = {\left[ {3\left( {4 - 4} \right)} \right]^{4 - 4}} = {0^0} \cr & {\text{This limit has the form }}{0^0}{\text{ }}\left( {{\text{See example 6, page 562}}} \right) \cr & {\text{Let }}y = \mathop {\lim }\limits_{x \to {4^ + }} {\left[ {3\left( {x - 4} \right)} \right]^{x - 4}} \cr & {\text{Take the natural log of each side}} \cr & \ln y = \ln \mathop {\lim }\limits_{x \to {4^ + }} \left[ {{3^{x - 4}}{{\left( {x - 4} \right)}^{x - 4}}} \right] \cr & {\text{Continuity}} \cr & \ln y = \mathop {\lim }\limits_{x \to {4^ + }} \left[ {\ln {3^{x - 4}}{{\left( {x - 4} \right)}^{x - 4}}} \right] \cr & {\text{Product property of logarithms}} \cr & \ln y = \mathop {\lim }\limits_{x \to {4^ + }} \left[ {\ln {3^{x - 4}} + \ln {{\left( {x - 4} \right)}^{x - 4}}} \right] \cr & \ln y = \mathop {\lim }\limits_{x \to {4^ + }} \left( {\ln {3^{x - 4}}} \right) + \mathop {\lim }\limits_{x \to {4^ + }} \left[ {\ln {{\left( {x - 4} \right)}^{x - 4}}} \right] \cr & \ln y = \mathop {\lim }\limits_{x \to {4^ + }} \left( {\left( {x - 4} \right)\ln 3} \right) + \mathop {\lim }\limits_{x \to {4^ + }} \left[ {\left( {x - 4} \right)\ln \left( {x - 4} \right)} \right] \cr & \ln y = \ln 3\mathop {\lim }\limits_{x \to {4^ + }} \left[ {\left( {x - 4} \right)} \right] + \mathop {\lim }\limits_{x \to {4^ + }} \left[ {\frac{{\ln \left( {x - 4} \right)}}{{1/\left( {x - 4} \right)}}} \right] \cr & {\text{Evaluating}} \cr & \ln y = 0 + \frac{\infty }{\infty } \cr & {\text{Using L'Hopital's rule}} \cr & \ln y = \ln 3\mathop {\lim }\limits_{x \to {4^ + }} \left[ {\left( {x - 4} \right)} \right] + \mathop {\lim }\limits_{x \to {4^ + }} \left[ {\frac{{\frac{d}{{dx}}\left[ {\ln \left( {x - 4} \right)} \right]}}{{\frac{d}{{dx}}\left[ {1/\left( {x - 4} \right)} \right]}}} \right] \cr & \ln y = \ln 3\mathop {\lim }\limits_{x \to {4^ + }} \left[ {\left( {x - 4} \right)} \right] + \mathop {\lim }\limits_{x \to {4^ + }} \frac{{\frac{1}{{x - 4}}}}{{ - \frac{1}{{{{\left( {x - 4} \right)}^2}}}}} \cr & \ln y = \ln 3\mathop {\lim }\limits_{x \to {4^ + }} \left[ {\left( {x - 4} \right)} \right] - \mathop {\lim }\limits_{x \to {4^ + }} \left( {x - 4} \right) \cr & {\text{Evaluating the limit}} \cr & \ln y = 0 \cr & y = 1 \cr & {\text{Therefore}} \cr & {\text{Let }}y = \mathop {\lim }\limits_{x \to {4^ + }} {\left[ {3\left( {x - 4} \right)} \right]^{x - 4}} \cr & \mathop {\lim }\limits_{x \to {4^ + }} {\left[ {3\left( {x - 4} \right)} \right]^{x - 4}} = 1 \cr} $$
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