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$$\lim\limits_{ (x,y,z) \to (0,0,0)} \frac {xyz^2}{x^2+y^4+z^6}$$

I checked that the limit does not exist but I cannot prove that.

I tried $y=mx$, $z=nx$ and also $y=x^m$, $z=x^n$ but they gave me nothing but the limit equals to zero.

Thanks a lot

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Let consider

$$\left|\frac {xyz^2}{x^2+y^4+z^6}\right|=\frac {|x||y|z^2}{x^2+y^4+z^6}$$

and let

$$\begin{cases} |x|=|X|\\\\ |y|=\sqrt{|Y|}\\\\ z=\sqrt[3] Z \end{cases}$$

then

$$\frac {|x||y|z^2}{x^2+y^4+z^6}=\frac {|X|\sqrt{|Y|}\sqrt[3] {Z^2}}{X^2+Y^2+Z^2}=\frac{\rho^{1+\frac12+\frac23}f(\theta,\phi)}{\rho^2}=\frac{\rho^{\frac{13}6}f(\theta,\phi)}{\rho^2}=\rho^\frac16f(\theta,\phi)\to 0$$

therefore since

$$\frac {|x||y|z^2}{x^2+y^4+z^6}\to 0 \implies \frac {xyz^2}{x^2+y^4+z^6}\to 0$$

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An always good way is to use spherical polar coordinates

$$\begin{cases} x = R\sin\theta\cos\phi \\ y = R\sin\theta\sin\phi \\ z = R\cos\phi \end{cases} $$

Which turns the limit into

$$\frac{R^4 \sin (\theta ) \cos (\theta ) \sin ^2(\phi ) \cos ^2(\phi )}{R^6 \cos ^6(\phi )+R^4 \sin ^4(\theta ) \sin ^4(\phi )+R^2 \cos ^2(\theta ) \sin ^2(\phi )}$$

Which goes to zero as $R\to 0$.

But following a linear path like

$$y = x^n$$ $$z = x^m$$

We get:

$$\frac{x^{2 m+n+1}}{x^{6 m}+x^{4 n}+x^2}$$

The limit of which does not exist.

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  • $\begingroup$ you should explain why it does not exist $\endgroup$ – user Feb 15 '18 at 9:26
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If $x=R\cos\theta,y^2=R\sin\theta\cos\phi,z^3=R\sin\theta\sin\phi$, then the expression is less than $R^{1/6}$.

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  • $\begingroup$ The idea is good but you need to better formalize it since you are considering a restriction for f(x,y,z). You can use the fact that $|f(x,y,z)|\to0$ and thus $f(x,y,z)\to 0$. $\endgroup$ – user Feb 15 '18 at 9:31

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