How can i solve


If I wanted to use polar coordinates, how can I convert the variables?

  • 1
    $\begingroup$ What is $z$ here? $\endgroup$ – Nosrati Sep 15 '18 at 13:43
  • $\begingroup$ Check for the z term, is it a parameter? $\endgroup$ – user Sep 15 '18 at 13:45
  • $\begingroup$ $z$ is a variable. I'm trying to prove the distinctness of the function $f(x,y,z)=\sqrt{(x-1)yz}$ in $(1,0,z)$. The increases are $h$, $k$ and $t$. $\endgroup$ – Marco Pittella Sep 15 '18 at 14:00

The limit does not exist. If you approach along a path with $h=0$ the quantity is always zero. If you approach along $h=k$ with $t=0$ the quantity is $\sqrt {\frac z2}$

  • $\begingroup$ Thanks for your answer. So, also with multivariable limits, i can use restrictions? $\endgroup$ – Marco Pittella Sep 15 '18 at 14:35
  • $\begingroup$ For multivariable limits the limit must be the same no matter how you approach the point of interest. One way to prove the limit does not exist is to find different paths that approach the point that lead to different limits. What I have really shown is that within any neighborhood of $(0,0,0)$ there are points where the expression is $0$ and other points where it is $\sqrt{\frac z2}$. The paths are just a nice way to show that. $\endgroup$ – Ross Millikan Sep 15 '18 at 15:02


Let use spherical coordinates with

  • $h=r\sin \phi \cos \theta$

  • $k=r\sin \phi \sin \theta$

  • $t=r\cos \phi$

to obtain

$$\frac{\sqrt{hk(z+t)}}{\sqrt{h^2+k^2+t^2}}=\sqrt {\sin^2 \phi \sin \theta\cos \theta(z+r\cos \phi)}$$

  • $\begingroup$ Thanks for your answer. Ok, then i know that $ |\sqrt{x}|=\sqrt{|x|} $, so i can write $|\sqrt{sin^2\phi sin\theta cos\theta(z+\rho cos\phi)}|=\sqrt{|sin^2\phi sin\theta cos\theta(z+\rho cos\phi)|}=\sqrt{|sin^2\phi|\cdot |sin\theta|\cdot|cos\theta|\cdot |z+\rho cos\phi|}\leq \sqrt{|z+\rho cos\phi|}$. $\endgroup$ – Marco Pittella Sep 15 '18 at 14:32
  • $\begingroup$ And now, i could apply the triangular inequality... $\endgroup$ – Marco Pittella Sep 15 '18 at 14:36
  • $\begingroup$ @MarcoPittella It seems that the limit doesn't exist unless $z \to 0$ too. $\endgroup$ – user Sep 15 '18 at 14:39

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