Calculus: Early Transcendentals (2nd Edition)

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

Chapter 12 - Functions of Several Veriables - 12.3 Limits and Continuity - 12.3 Exercises - Page 893: 16


$$\ln \left( 2 \right) + 1$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {{e^2},4} \right)} \ln \sqrt {xy} \cr & {\text{write }}\sqrt {xy} {\text{ as }}{\left( {xy} \right)^{1/2}} \cr & = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {{e^2},4} \right)} \ln {\left( {xy} \right)^{1/2}} \cr & {\text{logarithmic property}} \cr & = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {{e^2},4} \right)} \frac{1}{2}\ln \left( {xy} \right) \cr & {\text{law constant multiply}} \cr & = \frac{1}{2}\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {{e^2},4} \right)} \ln \left( {xy} \right) \cr & {\text{evaluate using the theorem 12}}{\text{.1}} \cr & = \frac{1}{2}\ln \left( {\left( {{e^2}} \right)4} \right) \cr & = \frac{1}{2}\ln \left( {4{e^2}} \right) \cr & = \frac{1}{2}\ln \left( 4 \right) + \frac{1}{2}\ln \left( {{e^2}} \right) \cr & = \ln \left( 2 \right) + 1 \cr} $$
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