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I want to know whether $\lim_{(x,y)\to (0,0)}\dfrac{x^2y^2}{x^3+y^3}$ exists or not. I tried to approximate to (0,0) from different "paths" and the result was always 0. For example,

$f(x,mx^2) = \dfrac{m^2x^3}{1+m^3x^3}$

But that doesn't show that the limit is 0.

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    See also: [Nonexistence of the limit $\lim_{(x,y)\rightarrow (0,0)} \frac{x^2y^2}{x^3+y^3}$](http://math.stackexchange.com/q/538770)2017-02-15

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Hint: There is real trouble if $x$ is the negative of $y$. If you don't want to take the easy way out and note that the function is not defined when $x=-y\ne 0$, let $(x,y)$ approach $(0,0)$ along the parametric path $x=t+t^2$, $y=-t+t^2$. If you want more dramatic behaviour, use the path $(t+t^3, -t+t^3)$.

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    @Thiago: More. The idea is that if we perturb $(t,-t)$ a little, then the "lead term" in $x^3+y^3$ will disappear, while the lead term in $x^2y^2$ won't. This will give us a good deal of control over the behaviour of the limit. Like you, I was changing to a one variable problem. Could have used $(t,-t+t^2)$, but have a liking for whatever symmetry might be around. Stuff like cancellation of lead terms happens a lot in one variable limits, particularly if we use series methods.2012-10-25
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Hint

Choose any $\alpha\ne0$. As $t\to-1$, consider the path $ (x,y)=\alpha(t^3+1)\left(\frac1{t^2},\frac1{t}\right)\to(0,0) $ For $\alpha=0$ use $ (x,y)=(t^3+1,0)\to(0,0) $ More Hint

For all $t\ne-1$, $\dfrac{x^2y^2}{x^3+y^3}=\alpha$.

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Hint: Use $\sqrt[3]{\frac{x^3+y^3}2}\ge \sqrt{xy}$.

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    Only when $x,y \geq 0$.2012-10-25