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

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.

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 $$x \to 0$$.

• Thanks a lot for the beautiful solution! One last thing - if the power of $y$ was $2$ in both the numerator and denominator, the limit would have been zero, right? i.e. $\lim_{(x,y)\to(0,0) }\frac{x^2 y^2}{x^6 + y^2}$ Apr 18, 2020 at 13:40
• @Dr.John You are welcome, and you are right. Replacing $y$ by $y^2$ would produce a limit of $0$. For $y\neq 0$ you would just estimate $0\leq \frac{x^2y^2}{x^6+y^2}\leq \frac{x^2y^2}{y^2}=x^2$. Apr 19, 2020 at 7:24
• Great. Thanks a lot again! Apr 19, 2020 at 12:53