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Today, I am asking to verify the continuity of the following multivariable function: $$f(x,y,z)=\begin{cases}\frac{xz-y^2}{x^2+y^2+z^2}&,\quad(x,y,z)\neq(0,0,0)\\{}\\0&,\quad(x,y,z)=(0,0,0)\end{cases}$$

The continuity can be easily rejected by seeing that the function has no limit at origin when we consider it on two paths: $$(x,0,0), ~~f\to 0 \quad \text{and} \quad (0,y,0), ~~f\to -1$$ As it is clear, this function is homogenous of degree zero also, so:

Can we say all the multivariable functions having the property above has no limit at the origin?

Indeed, we encounter many multivariable functions $\Bbb R^2\to\mathbb R$ or $\Bbb R^3\to\mathbb R$ and are asked to probe the continuity at the origin. I'd like you to light my mind about this point. Thanks for the time.

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Surely not - constant functions are homogeneous of degree zero as well – Cocopuffs Feb 22 '13 at 15:47
@Cocopuffs: So are we supposed to exclude constant function? In fact, I want to find a basic fast criteria for these functions and their limits at the origin. – Babak S. Feb 22 '13 at 15:50
Please do not delete questions with good answers. Others have devoted effort to answer your question; deleting the question is disrespectful of their effort and prevents others from benefiting from your question and its answers. – robjohn Mar 9 '13 at 10:05
@robjohn: Dear Rob, honestly, in that time; I didn't get any attentions to this question and thought maybe this one was not proper as it should be. So, I decided to remove it. I do know the points you kindly noted me. Thanks for remarking me that. I'll do that. Never happens again. ;-) – Babak S. Mar 9 '13 at 10:09
If $p(x,y,z)$ and $q(x,y,z)$ are both homogeneous of the same degree, then their quotient $f(x,y,z)=p(x,y,z)/q(x,y,z)$ is constant on any line through the origin, because for all real numbers $t$ we have $f(x_0,y_0,z_0)=f(tx_0,ty_0,tz_0).$ So for $f$ to have a limit at the origin, that constant has to be independent of the direction of the line. In other words, $f$ has to be a constant itself. If the denominator is homogeneous of a higher degree, then something will always break down, when you approach the origin... – Jyrki Lahtonen Mar 18 '13 at 17:56

Note that


are homogeneous functions (I'm not sure what you mean by "order" here), yet their limit when $\,(x,y,z)\to (0,0,0)\,$ exists, so I suppose one has to check each case separatedly.

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I meant $f(tx,ty,tz)=t^0f(x,y,z)$. – Babak S. Feb 22 '13 at 15:48
The limit of $f$ doesn't exist. Consider $f(x,0,0)$ as $x \to 0$. – Javier Feb 22 '13 at 16:02
You're right, Javier. Thanks. Edited – DonAntonio Feb 22 '13 at 17:39

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