limits - $displaystylemathop {lim }limits_{(x,y,z) to (0,0,0)} frac{{x{y^2}z}}{{{x^2} + 4{y^4} + 9{z^6}}}$

Consider the following limit $\displaystyle\mathop {\lim
}\limits_{(x,y,z) \to (0,0,0)} \frac{{x{y^2}z}}{{{x^2} + 4{y^4} +
9{z^6}}}$. If the limit does exist, compute its value; if it doesn't,
give a proof.

I feel that the limit does exist, judging by the exponents of $z$. So as usual when it comes to showing the existence of a three-variable limit, I tried to convert to spherical coordinates. But that method can't seem to work in this case. So is there any way to evaluate its limit? Or the limit does not exist at all and I'd better try to disprove its existence?
Any help is much appreciated. Thanks!

Answer

This is close to Christian Blatter's answer, but I'll avoid the trig functions and spell out some more details. For any $x,y,z$, define $M$ to be the maximum of $x^2$, $y^4$, and $z^6$. Notice that $M>0$ for all $(x,y,z)\neq(0,0,0)$ and that $M\to0$ as $(x,y,z)\to(0,0,0)$.

Consider the denominator $x^2+4y^4+9z^6$ of the fraction in the problem. One of the three summands here is $M$, possibly multiplied by 4 or 9, and the other two summands are nonnegative. So the whole denominator is $\geq M$.

Next consider the numerator, $xy^2z$. By definition of $M$, we have $|x|\leq M^{1/2}$, $|y|\leq M^{1/4}$ (so $y^2\leq M^{1/2}$), and $z\leq M^{1/6}$. Thus, the absolute value of the numerator is $\leq M^{7/6}$.

Combining the lower bound on the denominator and the upper bound on the absolute value of the numerator, we get that the whole fraction is bounded in absolute value by $M^{1/6}$. Since $M$ approaches $0$ as $(x,y,z)\to(0,0,0)$, so does the fraction.