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Find the limit at the origin: $$f(x,y)={\frac{x^3+y^2}{x^2+y}}$$

I already tried all the methods for proving that the limit is zero at the origin but had no success.

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    @anon where the idea came from of take $x=t$ e $y=−t^2+h$ ?2012-11-26
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    The idea is to perturb the path of singularities $(t,-t^2)$ ever so slightly. It is easier to perturb it in the $y$-coordinate because of how the denominator works out.2012-11-26
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    @anon then this method is common?2012-11-26

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For the function $$f(x,y)=\frac{x^3+y^2}{x^2+y},$$ note that we have $$f(t,-t^2+h)=\frac{t^4+t^3-2th+h^2}{h}=\frac{t^4+t^3}{h}-2t+h.$$ Now provided $t,h$ go separately to $0$ in any way, the values $x=t,y=-t^2+h$ will each approach $0$. So if the limit existed, the fraction $(t^4+t^3)/h$ would have to be bounded as $t,h \to 0$. But this doesn't hold since if we take say $h=t^4$, then $$(t^4+t^3)/h=(t^4+t^3)/t^4=1+1/t.$$

More simply: suppose the limit exists and is $a$. Then the one variable limit $$\lim_{t \to 0} f(t,-t^2+t^4)$$ would also be $a$. However $$f(t,-t^2+t^4)=t^4-2t^2+1+1/t,$$ which as $t \to 0$ has no limit (approaching $\infty$ or $-\infty$ depending on whether $t$ approaches $0$ from above or below).

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The previous idea I wrote is wrong since we cannot multiply $\min\{x,y\}\le y\le\max\{x,y\}$ by $y$ and retain the direction of the inequalities when $y<0$.

It is clear that the path to the origin on the $x$-axis results in a limit of zero, since plugging in zero for $y$ in the expression results in $f(x,0)=x$. However this is only useful if the limit exists.

Observe that the curve $\ell$ of $y=-x^2$ traces out a path to $(0,0)$ and yet the function isn't defined on this path (as it would mean division by zero), and indeed $f$ takes on arbitrarily large values around the curve $\ell$. This path goes through any disc around the origin, and thus given any disc we can put down a curve $\gamma$ contained in the disc that crosses over $\ell$; the function $f$ restricted to $\gamma$ will have a singularity when it intersects $\ell$, and $f$ will be unbounded on this path so that $f$ is unbounded on the disc around the origin, no matter how small we chose this disc to be. The limit cannot exist under these circumstances.

This isn't very formal and detailed and would probably not be good to put down for homework; its only utility is that it is relatively straightforward and persuasive for the idea the limit doesn't exist.

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    Dear anon, how about taking max$\{|x|,|y| \}$ instead? Isn't it better?2012-11-22
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    @BabakSorouh How do you mean?2012-11-22
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    Because, facing $\sqrt{x^2+y^2}<\delta$ such this case, always makes me to take $|x|<\delta$ and $|y|<\delta$. And I think you are using this fact to show that the limit is zero. In fact you are using max$\{|x|,|y|\}$.2012-11-22
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    @BabakSorouh I was thinking conceptually in terms of squeezing, actually. It is clear that both min and max tend to zero, and it should be easy to see and prove that a function bounded between them will have the same limit where the upper/lower bounds are equal. But whatever floats your boat.2012-11-22
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    Right. Thanks for your patience. :)2012-11-22
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    The inequality using min and max of x,y seems to me to require one to assume x,y positive in order that the inequalities do not reverse.2012-11-22
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    @coffeemath Right, this is silly.2012-11-23
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    @CarlosHenriqueFelicio You should stop by and check what I've written now. I recommend looking at coffeemath's answer and accepting it if/when you understand it.2012-11-23
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    anon: Yes. What struck me with the function is that the top is bounded near zero and is not a multiple of the bottom, which along $y=-x^2$ is zero. I think in all such cases the limit blows up.2012-11-23