A limit question of 3-variable-functions. $\lim\limits_{ (x,y,z) \to (0,0,0)} \frac {xyz^2}{x^2+y^4+z^6}$

Question

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

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

Answer 1

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

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

Answer 2

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 $R^{1/6}$.

Answer 3

We have $|x| \le \sqrt{x^2 + y^4 + z^6}$ and $|y| \le \sqrt[4]{x^2 + y^4 + z^6}$ and $|z| \le \sqrt[6]{x^2 + y^4 + z^6}$, so we got:

$$\left| \frac{x y z^2}{x^2 + y^4 + z^6} \right| \le (x^2 + y^4 + z^6)^{\frac{1}{2} + \frac{1}{4} + \frac{2}{6} - 1} = (x^2 + y^4 + z^6)^{\frac{1}{12}}$$

Since the right has limit $0$, the limit converges to $0$.