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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 $y=−t^2+h$ ? –  Henfe Nov 26 '12 at 21:51
    
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. –  anon Nov 26 '12 at 21:53
    
@anon then this method is common? –  Henfe Nov 26 '12 at 22:21
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2 Answers

up vote 2 down vote accepted

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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Thank you for your attention. –  Henfe Nov 23 '12 at 12:07
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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? –  B. S. Nov 22 '12 at 19:07
    
@BabakSorouh How do you mean? –  anon Nov 22 '12 at 19:10
    
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$\{|x|,|y|\}$. –  B. S. Nov 22 '12 at 19:16
    
@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. –  anon Nov 22 '12 at 19:21
    
Right. Thanks for your patience. :) –  B. S. Nov 22 '12 at 19:23
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