<s>[INST] Prove $\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0$ without $\varepsilon - \delta$. Unlike Multivariable Delta Epsilon Proof $\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4}$ --- looking for a hint I would like to avoid the $\varepsilon - \delta$ criterium.

Prove $$\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0 \,.$$

Approaching this limit from $y=0$, $x=0$, $y=x$, $y=x^2$ etcetera all yields 0 as value, so my proposal is that this limit is indeed 0.
I have been able to solve most similar limits so far by finding some convergent upper bound for the absolute limit, but with this one the difference between the numerator and the denominator is so small I can't find anything to fit inbetween. For example, $(x,y)\to(0,0)$,
$$ \left| \frac{x^2 y}{x^2 + y^2} \right| \le \left| \frac{(x^2 + y^2)y}{x^2 + y^2} \right| \to 0 \,. $$
Also, Continuity of $\frac{x^3y^2}{x^4+y^4}$ at $(0,0)$? and Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous. contain some helpful hints.
 [/INST] Hint 1: $$\frac{x^2 y^3}{x^4+y^4}=\frac{y}{(x/y)^2+(y/x)^2}. $$ 
Hint 2: $f(u)=u^2+u^{-2}$ has an absolute minimum.
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