Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.2 Limits and Continuity - 14.2 Exercises - Page 950: 20


limit does not exist.

Work Step by Step

Given: $\lim\limits_{(x,y,z) \to (0,0,0)}\frac{xy+yz}{ {x^{2}+y^{2}+z^{2}}}$ We notice that if we directly substitute limits in the given function $f(x,y)=\frac{xy+yz}{ {x^{2}+y^{2}+z^{2}}}$ Then $f(0,0)=\frac{0}{0}$ We get intermediate form. Therefore, we will calculate limit of function in following way. To evaluate limit along x-axis; put $y=0,z=0$ $f(x,0,0)=\frac{y(x+z)}{ {x^{2}+y^{2}+z^{2}}}=\frac{0(x+0)}{ {x^{2}+0^{2}+0^{2}}}=0$ To evaluate limit along x-axis; put $x=0,z=0$ $f(0,y,0)=\frac{y(x+z)}{ {x^{2}+y^{2}+z^{2}}}=\frac{y(0+0)}{ {0^{2}+y^{2}+0^{2}}}=0$ To evaluate limit along x-axis; put $x=0,y=0$ $f(0,0,z)=\frac{y(x+z)}{ {x^{2}+y^{2}+z^{2}}}=\frac{0(0+z)}{ {0^{2}+0^{2}+z^{2}}}=0$ Although the obtained identical limits along the axes do not show that the given limit is 0. Then approach along another curve let us say $x=y=z$ , $z\ne 0$ for $y\ne 0$ $f(x,x,x)=\frac{x.x+x.x}{ {x^{2}+x^{2}+x^{2}}}=\frac{2}{3}$ Thus , the limit along x-axis does not coincide with the limit along the curve $x=y=z$. For a limit to exist, all the paths must converge to the same point.Since, function $f(x,y,z)$ has two different values along two different paths, it follows that limit does not exist, Hence, the limit does not exist.
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