Limit as $(x,y)\to(0,0)$ for $\frac{x^3y^2}{x^4+y^6}$

Question

I want to show that the limit $(x,y)\to(0,0)$ is $0$ for $$\frac{x^3y^2}{x^4+y^6}$$

I wrote it as $$\frac{\frac{x}{y}x^2y^3}{(x^2)^2+(y^3)^2}$$ We know $x^2<\sqrt{(x^2)^2+(y^3)^2}$ and $y^3<\sqrt{(x^2)^2+(y^3)^2}$.

Therefore, it seems like it is of the form $\frac{x}{y}(\text{bounded function})$. But clearly, $\frac{x}{y}$ is not bounded, and so we can't say $\lim_{(x,y)\to(0,0)}$ is $0$. However, Wolfram Alpha says that the limit will be $0$. Where am I going wrong?

Answer 1

Courtesy of Lord Shark the Unknown.

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More generally, consider

$$\lim_{(x,y)\to(0,0)}\frac{|x|^a|y|^b}{|x|^c+|y|^d}$$

Now define $u=|x|^c+|y|^d$ and see that

$$u\ge|x|^c\implies u^{a/c}\ge|x|^a\\u\ge|y|^d\implies u^{b/d}\ge|y|^b$$

Thus

$$0\le\frac{|x|^a|y|^b}{|x|^c+|y|^d}\le u^{\frac ac+\frac bd-1}$$

So if

$$\frac ac+\frac bd-1>0$$

Then the limit must be zero.