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: 56

Answer

$$0$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{x - \sqrt {xz} - \sqrt {xy} + \sqrt {yz} }}{{x - \sqrt {xz} + \sqrt {xy} - \sqrt {yz} }} \cr & {\text{Evaluating}} \cr & \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{x - \sqrt {xz} - \sqrt {xy} + \sqrt {yz} }}{{x - \sqrt {xz} + \sqrt {xy} - \sqrt {yz} }} = \frac{{1 - \sqrt {\left( 1 \right)\left( 1 \right)} - \sqrt {\left( 1 \right)\left( 1 \right)} + \sqrt {\left( 1 \right)\left( 1 \right)} }}{{1 - \sqrt {\left( 1 \right)\left( 1 \right)} + \sqrt {\left( 1 \right)\left( 1 \right)} - \sqrt {\left( 1 \right)\left( 1 \right)} }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1 - 1 - 1 + 1}}{{1 - 1 + 1 - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{0}{0}{\text{Ind}} \cr & {\text{Write the function as}} \cr & = \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{x - \sqrt x \sqrt z - \sqrt x \sqrt y + \sqrt y \sqrt z }}{{x - \sqrt x \sqrt z + \sqrt x \sqrt y - \sqrt y \sqrt z }} \cr & {\text{Factoring by grouping terms}} \cr & = \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{\sqrt x \left( {\sqrt x - \sqrt z } \right) - \sqrt y \left( {\sqrt x - \sqrt z } \right)}}{{\sqrt x \left( {\sqrt x - \sqrt z } \right) + \sqrt y \left( {\sqrt x - \sqrt z } \right)}} \cr & = \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{\left( {\sqrt x - \sqrt z } \right)\left( {\sqrt x - \sqrt y } \right)}}{{\left( {\sqrt x - \sqrt z } \right)\left( {\sqrt x + \sqrt y } \right)}} \cr & {\text{Simplifying}} \cr & = \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{\sqrt x - \sqrt y }}{{\sqrt x + \sqrt y }} \cr & {\text{Evaluating}} \cr & \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{\sqrt x - \sqrt y }}{{\sqrt x + \sqrt y }} = \frac{{\sqrt 1 - \sqrt 1 }}{{\sqrt 1 + \sqrt 1 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{0}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0 \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.