Show $\lim_{(x,y) \to (0,0)}\frac{3 x^3 y^2 - x y^6}{(x^2 + y^4)^2}= 0$

Question

When trying to verify the Cauchy-Riemann equations in $z = 0$ for $f(x + i y) = \frac{x y^2 (x + i y)}{x^2 + y^4}$ for $x + i y \ne 0$ and $0$ for $x + i y = 0$, I encountered the limit $$\lim_{(x,y) \to (0,0)}\frac{3 x^3 y^2 - x y^6}{(x^2 + y^4)^2}$$ as I decomposed $\frac{x y^2 (x + i y)}{x^2 + y^4} = \frac{x^2 y^2}{x^2 + y^4} + i \frac{x y^3}{x^2 + y^4} = u(x,y) + i v(x,y)$. Then $\frac{\partial v(x,y)}{\partial x} = \frac{3 x^3 y^2 - x y^6}{(x^2 + y^4)^2}$. From plotting I suspect that the limit is zero, however I have't managed to prove it.

It tried using $u := x^2 + y^4$ as $u \to 0$ $\iff$ $(x,y) \to (0,0)$, which yields $| x | \le x^{1/2}$ and $| y | \le u^{1/4}$. Thus $$ \left| \frac{3 x^3 y^2 - x y^6}{(x^2 + y^4)^2} \right| \overset{\triangle \ne}{\le} \frac{3 | x |^3 | y |^2 + | x | | y |^6}{(x^2 + y^4)^2} \le \frac{3 u^{3/2} u^{1/2} + u^{1/2} u ^{6/4}}{u^2} = 4 \ne 0. $$ I also tried polar coordinates $x = r \cos(\theta)$, $y = r \sin(\theta)$ with $r > 0$ and $\theta \in [0, 2 \pi)$: $$ \frac{3 x^3 y^2 - x y^6}{(x^2 + y^4)^2} = \frac{3 r^5 \cos(\theta)^3 \sin(\theta)^2 - r^7 \cos(\theta) \sin^6(\theta)}{(r^2 \cos^2(\theta) + r^4 \sin^4(\theta))^2} = \frac{3 r \cos(\theta)^3 \sin(\theta)^2 - r^3 \cos(\theta) \sin^6(\theta)}{(\cos^2(\theta) + r^2 \sin^4(\theta))^2} $$ But I don't know how to simplify this further.

Answer 1

$\lim_{(x,y) \to (0,0)}\frac{3 x^3 y^2 - x y^6}{(x^2 + y^4)^2}$ does not exist:

let $h(x,y):=\frac{3 x^3 y^2 - x y^6}{(x^2 + y^4)^2}$ for $(x,y) \ne (0,0).$

Then $h(x,x) \to 0$ as $x \to 0$ and $h(y^2,y) \to 1/2$ as $y \to 0.$

Answer 2

It may be interesting to substitute $u = y^2$ and consider the limit at $u \to 0^+$, so we get $$ \lim_{(x,y) \to (0,0)}\frac{3 x^3 y^2 - x y^6}{(x^2 + y^4)^2} = \lim_{(x,u) \to (0,0^+)}\frac{3 x^3 u - x u^3}{(x^2 + u^2)^2} = \lim_{(x,u) \to (0,0^+)}\frac{xu(3 x^2 - u^2)}{(x^2 + u^2)^2} $$ and now polar coordinates really help.