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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 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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    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