Calculating (multivariable) limit using squeeze thm. or polar coordinates ($\lim_{(x,y)\to(0,0) }\frac{x^2 y}{x^6 + y}$)

Question

I need to calculate the following limit. $$\lim_{(x,y)\to(0,0) }\frac{x^2 y}{x^6 + y}$$ and the recommended hint is to use the squeeze them, or polar coordinates. My problem is that I am not sure how to bound the denominator from below in a meaningful way. For example, $x^6+y\geq y$, but this does not help me with the calculation. In polar coordinates, $$ f(r \cos\theta , r\sin \theta) = \frac{r^3 \cos^2 \theta \sin \theta }{r^6 \cos^6 \theta + r\sin \theta } = \frac{r^2 \cos^2 \theta \sin \theta }{r^5 \cos^6 \theta + \sin \theta } $$ and I am also not sure how to deal with the $\sin \theta $ term in the denominator.

I have also tried using the paths $y=x,y=x^2,y=x^4,y=x^6$ and they all yield a limit of zero, but I want to prove the result in general.

Thanks in advance

Answer 1

If you rewrite the expression for $y \neq 0$ as

$$\frac{x^2 y}{x^6 + y} = \frac{x^2}{\frac{x^6}{y}+1}$$

you may guess that a good try to show divergence is to choose

$$y= \frac{x^6}{-1+x^3}\stackrel{x\to 0}{\longrightarrow}0$$

Indeed, you get

$$\frac{x^2}{\frac{x^6}{y}+1} = \frac{x^2}{x^3} =\frac 1{x}$$

which is obviously divergent for $x \to 0$.