Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.2 - Limits and Continuity - 14.2 Exercise - Page 910: 13

Answer

Given: $f(x,y)=\frac{xy}{\sqrt {x^{2}+y^{2}}}$ Then $f(0,0)=\frac{0}{0}$ To evaluate limit along the x-axis; put $y=0$ $f(x,0)=\frac{xy}{\sqrt {x^{2}+y^{2}}}=0$ To evaluate limit along y-axis; put $x=1$ $f(1,y)=\frac{xy}{\sqrt {x^{2}+y^{2}}}=0$ However, the obtained identical limits along the axes do not show that the given limit is 0. Then, approach (0,0) along another line, $y=x$ and $x≠0$ $f(x,x)=\frac{xy}{\sqrt {x^{2}+y^{2}}}=\frac{x^{2}}{\sqrt {x^{2}+x^{2}}}=\frac{x^{2}}{\sqrt {2x^{2}}}=\frac{x}{\sqrt 2}$ For a limit to exist, all the paths must converge to the same point. Since f(x,y) has two different values along two different paths, it follows that the limit does not exist.

Work Step by Step

By direct substitution: $f(0,0)=\frac{0}{sqrt(0)}$ is indeterminate Along the x-axis; y=0 $f(x,0)=\frac{0}{sqrt(x^{2})}=0$ Along the y-axis;x=0 $f(0,y)=\frac{0}{sqrt(y^{2})}=0$ Therefore the limit exists
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.