# $\lim _{(x,y) \to (0,0)}$ $\frac{(x^3-y^2)^2}{x^4+y^4}$- 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.

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Can you detail the inequality? – Davide Giraudo Feb 1 '12 at 12:14
Sure. I tried to increase the numerator and decrease the denominator as I wrote. What other details would you like? – Jozef Feb 1 '12 at 12:20
It seems wrong, since if we take $x=0$ in the LHS, we have $1$ and $0$ in the RHS. – Davide Giraudo Feb 1 '12 at 12:21
$(x^3)^2$ is not always larger than $(x^3-y^2)^2$. – Chris Eagle Feb 1 '12 at 12:23

This can be done simply:

Take a path with $x=0$ and $y\rightarrow0$; then ${(x^3-y^2)^2\over x^4+y^4}={y^4\over y^4}=1$.

Take a path with $y=0$ and $x\rightarrow0$; then ${(x^3-y^2)^2\over x^4+y^4}={x^6\over x^4}=x^2\rightarrow0$.

So $\lim\limits_{(x,y)\rightarrow(0,0)} { (x^3-y^2)^2 \over x^4+y^4}$ does not exist.

Your second argument is correct. For $y=Kx$, you have: $${(x^3-y^2)^2\over x^4+y^4} ={(x^3-K^2x^2)^2\over x^4+K^4 x^4} ={(x-K^2)^2\over 1+K^4}\quad\buildrel{x\rightarrow0}\over{\longrightarrow}\quad{K^4\over 1+K^4} .$$ From this, you can see that the limit does not exist (take, e.g., $K=0$ and $K=1$).
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 $x,y\neq 0$.