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I want to determine $$\lim\limits_{(x,y) \to (0,0)} \frac{x^2·y^3}{x^4+y^6}$$

I'm sure the limit exists, it's zero because I tried to find other different limits in line and parabola points ($(x,mx)$ and $(x,mx^2)$) but every time I got zero. So, I have to use the definition or sequences to confirm.

Martin Sleziak
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Aarón David Arroyo Torres
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  • It may be helpful for you to look at similar questions, see for example http://math.stackexchange.com/questions/174190/prove-that-lim-limits-x-y-to-0-0-fracxy2x2-y4-0 or http://math.stackexchange.com/questions/346088/find-lim-limits-x-y-to0-0-fracxy2-x2-y4?lq=1 – jim Feb 13 '16 at 21:18

2 Answers2

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There is no limit. Along the semicubical parabola $x=t^3$, $y=t^2$ the limit is equal to 1/2. S.G.

user312865
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Note that if you replace $x^2$ by $x'$ and $y^3$ by $y'$, then the expression reduces to $\frac{x'y'}{x'^2 + y'^2}$.

If you now take $x' = y'$, then you get $1/2$. This shows that the limit does in fact not exist, as you found correctly that some approaches lead to $0$.

If you do not like the replacement: when you approach via $(t^3, t^2)$ you get $1/2$.

quid
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