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I need to calculate the limit of the function: $ f(x,y) = \dfrac{2xy^3}{3x^2 + y^6 } $ at $(x,y)=(0,0)$.

When I check the orbits $y=x , y=x^{1/3} $ , I can see that such a limit doesn't exist. But if this limit doesn't exist, I should also see it in polar coordinates (I'll see that the limit is dependent on $\theta$ ). In this case I get that this limit is always $0$.

Where is my misunderstanding? Can someone help me see that the limit in polar coordinates does depend on $\theta$?

Thanks everyone!

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In the denominator do you mean $y^2$ (or $y^3$)? –  user39572 Oct 4 '12 at 19:08
    
Thanks ! I edited my typo! –  joshua Oct 4 '12 at 19:11

1 Answer 1

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Actually, no, if you fix $\theta$ and let $r \rightarrow 0^+$ the ratio you indicate does indeed go to $0.$ It is just that this still does not mean there is an actual limit. Probably it is worth emphasizing that the rate at which the thing goes to $0$ varies with $\theta,$ in such a way that if you take $x = y^3$ the value of the ratio is always $1/2,$ while if you take $x = - y^3$ the ratio is always $-1/2.$

Put another way, if I let $r \rightarrow 0^+$ but demand $ \theta $ vary, in the first quadrant, in such a way that $r^2 \sin^3 \theta = \cos \theta,$ I would get the value $1/2.$

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Thanks... Just wanted to verify that there's no other way. Thanks a lot ! –  joshua Oct 4 '12 at 22:51

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