<s>[INST] $\lim _{(x,y) \to (0,0)}$ $\frac{(x^3-y^2)^2}{x^4+y^4}$- two methods different conclusion. I tried to use couple of inequalities in this way:
$0\leq\left|\frac{(x^3-y^2)^2}{x^4+y^4}\right| \leq \frac{(x^3)^2}{x^4}=x^2$, so now I can use the squeeze theorem and  conclude that this limit is 0, but when I choose a route such as $y=Kx$ I get that the limit is $\frac {K^4}{1+K^4}$, which depends on K, so apparently there's no limit when $(x,y) \to (0,0)$. 
Where's my mistake?
Thanks a lot.
 [/INST] The second result is the correct one. To see this, write $\frac{(x^3-y^2)^2}{x^4+y^4}=\frac{x^6}{x^4+y^4}-2x\frac{x^2y^2}{x^4+y^4}+\frac{y^4}{x^4+y^4}$. Since 
$$\left|\frac{x^6}{x^4+y^4}-2x\frac{x^2y^2}{x^4+y^4}\right|\leq x^2+|x|\frac{x^4+y^4}{x^4+y^4}=x^2+|x|,$$
we just have to study $\lim_{(x,y)\to (0,0)}\frac{y^4}{x^4+y^4}$. It doesn't exist since putting $g(x,y)= \frac{y^4}{x^4+y^4}$ we have $g(x,0)=0$ and $g(0,y)=0$ for $x,y\neq 0$.
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