Is the function $h(x,y)=\frac{x^3y^3}{x^4+y^6}$ if $(x,y)=(0,0)$ ; $h(0,0) = 0$ continuous at $(0,0)$?

Question

I think it is, because I cannot find a set such that the limit through it is different from $0$. However, polar coordinates don't simplify the way and I'm not able to bound the function. Any hints on how to proceed?

Answer 1

The function is continuous at $(0,0)$. It is enough to consider $x,y>0$. [Because $(|x|,|y|) \to 0$ as $(x,y) \to 0$ and $|h(x,y)|=\frac {|x|^{3} |y|^{3}} {|x|^{4}+|y|^{6}}$]. Since $(x^{2}-y^{3})^{2} \geq 0$ we get $x^{4}+y^{6} \geq 2x^{2}y^{3}$ so $ 0<\frac {x^{3}y^{3}} {x^{4}+y^{6}} \leq\frac {x^{3}y^{3}} {2x^{2}y{3}}=\frac x 2$ and $\frac x 2$ tends to $0$.

Answer 2

By $\text{HM}\le\text{GM}$,

$\dfrac{2}{\frac{1}{x^4}+\frac{1}{y^6}}\le\sqrt{x^4\cdot y^6}=x^2\cdot y^3\implies\dfrac{x^4y^6}{x^4+y^6}\le\dfrac{x^2\cdot y^3}{2}\implies h(x,y)=\dfrac{x^3y^3}{x^4+y^6}\le\dfrac{x}{2}\to0\text{ as }x\to0$

Hence $h(x,y)$ is continuous at $(0,0).$