Calculus 10th Edition

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

Chapter 13 - Functions of Several Variables - 13.2 Exercises - Page 887: 21

Answer

The limit is $$\lim_{(x,y,z)\to(1,3,4)}\sqrt{x+y+z}=2\sqrt2.$$ The function is continuous everywhere in its' domain.

Work Step by Step

We will calculate the limit by simple substitution: $$\lim_{(x,y,z)\to(1,3,4)}\sqrt{x+y+z}=\sqrt{1+3+4}=\sqrt{8}=2\sqrt2.$$ This function is continuous everywhere at its' domain because we can always perform the substitution and get $$\lim_{(x,y,z)\to(x_0,y_0,z_0)}\sqrt{x+y+z}=\sqrt{x_0+y_0+z_0},$$ where the ordered triple $(x_0,y_0,z_0)$ is from the domain.
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