## Calculus: Early Transcendentals 8th Edition

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 $f(0,0)=\frac{0}{0}$ Therefore, we will calculate the limit of the function in the 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 $x=0,y=0$ $f(0,0,z)=\frac{y(x+z)}{ {x^{2}+y^{2}+z^{2}}}=\frac{0(0+z)}{ {0^{2}+0^{2}+z^{2}}}=0$ Then approach along another curve, $x=y=z$, $z\ne 0$ 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 the 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. Hence, the limit does not exist.