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Does the following limit exist?

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

I tried to solve this problem using polar coordinates, but I can't simplify it. I tried the squeeze theorem, I got $0.5$, but I think this is incorrect.

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Let $x^2=y^3$ therefore we have $$\lim\limits_{(x,y)\to(0,0)}\frac{x^2y^3}{x^4+(x^2+y^3)^2}=\lim\limits_{x\to0}\frac{x^4}{5x^4}=\frac{1}{5} \ \ \ (1)$$ and if we consider $x=y$ then $$\lim\limits_{(x,y)\to(0,0)}\frac{x^2y^3}{x^4+(x^2+y^3)^2}=0 \ \ \ \ (2)$$ so that from (1) and (2) we see that limit doesn't exist.

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Hint: Consider the family of paths $y = kx^{\frac{2}{3}}$.

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Oh i didnt know that we can use to a power of a fraction. – ys wong Aug 17 '14 at 15:21
What if the limit actually exist? – ys wong Aug 17 '14 at 15:22
If the limit exists, the limit along any path exists and the limits along any two paths is the same. Note that using paths you can only show that a two-variable limit does not exist, you can't use paths to prove that a two-variable limit does exist. – Michael Albanese Aug 17 '14 at 15:40

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