<s>[INST] Limit at (0,0) of $\frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4}$ $$\text{Let}\ f(x,y) = \frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4}$$
WolframAlpha tells me that $\lim_{(x,y) \to (0,0)} f(x,y)$ does not exist. To prove the non-existence of the limit, I tried three different paths ($y=kx, y=kx^2, y=kx^3)$ and they all equal zero. From the graph of the function it looks like the limit is indeed zero. Also I thought I was able to prove the limit does indeed exist at $(0,0)$ with the epsilon-delta definition:
$$y^2 \leq x^2+y^2 \leq x^2 + y^4 \iff \frac{y^2}{x^2+y^4} \leq 1 \overset{|x|<1}{\iff} \frac{|x|y^2}{x^2+y^4} \leq 1 \iff \frac{|x|y^2\sqrt{x^2+y^2}}{x^2+y^4} \leq \sqrt{x^2+y^2} \iff \left|\frac{xy^2\sqrt{x^2+y^2}}{x^2+y^4} - 0\right|\leq \left|\sqrt{(x-0)^2+(y-0)^2}\right| \lt δ := ε $$
So for $δ = ε$ we have that $\|x -(0,0)\| < δ \implies |f(x,y) - 0| < ε$
I'm confused
 [/INST] As an alternative, by $y^2=v$ and polar coordinates, with $x=\rho\cos \theta$ and $v=\rho|\sin \theta|$, we obtain
$$ \frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4} =\frac{xv\sqrt{x^2+v}}{x^2 +v^2}=\sqrt \rho \cos \theta |\sin \theta| \sqrt{\rho\cos^2 \theta|\sin \theta|}\to 0$$
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