\(\lim_{(x,y,z)\rightarrow (0,0,0)}\frac{{x}^{2}{y}^{2}{z}^{2}}{{x}^{2}{y}^{2}{z}^{2}+(\left | x \right | +\left | y \right |+\left | z \right |)^2}\)  


I think it is 

\(\lim_{(x,y,z)\rightarrow (0,0,0)}\frac{1}{1+\frac{{(\left | x \right | +\left | y \right |+\left | z \right |)^2}}{x^2y^2Z^2}}\)


With homogeneous function in the fraction easy we can proof the degree of homogeneity up is 2 and down 6 so the lim -> infinity but i cant proove it.

Thank you!

 Nov 13, 2019

You almost have the answer. The fraction in the denominator of the complex fraction is a ratio of two homogeneous functions; the top has degree two and the bottom has degree 6. So the fraction has degree -6+2 =-4, so its limit is undefined (not infinite) and that makes the limit of your original expression undefined.


For a more rigorous way of showing the limit is undefined convert the function\(F(x, y,z)=\frac{(|x|+|y|+|z|)^2}{x^2y^2z^2}\)into spherical coordinates and take the limit as r goes to zero. When converted you would have  \(F= \frac{1}{r^4}g(\theta,\phi)=r^{-4}g(\theta,\phi)\), where \(g\) is continuous on the closed interval \([0,2\pi]\) and therefore bounded from both above and below and is independent of r. This means that the limit is not defined (see Vipul Naik's monogram "LIMITS IN MULTIVARIABLE CALCULUS" that can be downloaded at



 Nov 14, 2019

unfortunately I need to retract what I said about conversion to spherical coordinate since the resulting function \(g\) is a function of two variables and it makes no sense to say that it is continuous on \([0,2\pi]\); on \([0,2\pi]\times[0, 2\pi] \)maybe, but not on a single closed interval. Even then the function may not be continuous since in the denominator of \(g\) both \(sin(\theta)\)and \(sin(\phi)\)appear as factors and the function is not defined at \((0,0)\). I tried to extend Vipul Naik's two dimensional claims to three dimensions and it did not quite work the same way. I will keep working on this since I am definitely interested in it myself.

 Nov 14, 2019

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