How to prove that $\lim_{(x,y)\to (0,0)} \frac{xy^5}{x^4+y^6}=0$

Question

How can I apply the squeeze theorem to show $$\lim_{(x,y)\to (0,0)} \frac{xy^5}{x^4+y^6}=0$$

Answer 1

Hint: for $xy = 0$ the expression is zero, for $xy\ne 0$ bound the denominator using the weighted AGM inequality with weights 1 and 3: $$|x||y|^{9/2}\le\frac{x^4 + 3y^6}4\le x^4 + y^6.$$

Answer 2

If $|x|\geqslant|y|^{4/3}$ and $x\ne0$, $$\left|\frac{xy^5}{x^4+y^6}\right|\leqslant\left|\frac{xy^5}{x^4}\right|=\frac{|y|^5}{|x|^3}\leqslant|x|^{3/4}$$ If $|x|\leqslant|y|^{4/3}$ and $y\ne0$, $$\left|\frac{xy^5}{x^4+y^6}\right|\leqslant\left|\frac{xy^5}{y^6}\right|=\frac{|x|}{|y|}\leqslant|y|^{1/3}$$ Thus, for every $(x,y)\ne(0,0)$, $$\left|\frac{xy^5}{x^4+y^6}\right|\leqslant|x|^{3/4}+|y|^{1/3}$$ The limit of the LHS when $(x,y)\to(0,0)$ follows.