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Im trying to calculate (or prove that the limit dosent exists)

$ \lim_{\left(x,y\right)\to\left(0,0\right)}\frac{xy^{2}}{x^{4}+y^{2}} $

Wolfram says the limit does not exists, but for every path I choose the limit ends up being 0.

If someone can tell if the limit exists or not, it would be helpful.

Thanks in advance

FreeZe
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  • Try $y=x^t$ for some adequate value of $t$... or try to find this ADEQUATE value. – Tito Eliatron Oct 30 '20 at 18:29
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    $\left| \frac{xy^{2}}{x^{4}+y^{2}} \right| \le \left| \frac{xy^{2}}{y^{2}} \right| = |x|$, so ... – Martin R Oct 30 '20 at 18:31
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    See https://math.stackexchange.com/q/66226/42969 for a general result on this type of limit. – Martin R Oct 30 '20 at 18:34
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    You may have miscopied this. The usual would be $\frac{x^2 y}{x^4 + y^2} $ – Will Jagy Oct 30 '20 at 19:47
  • $$\left|\frac{xy^2}{x^4+y^2}\right|\le \frac{|x|(x^4+y^2)}{x^4+y^2}=|x|$$ – Mark Viola Oct 30 '20 at 20:03
  • @Will Jagy are we obligated to solve just the usual well known problems? – FreeZe Oct 30 '20 at 21:54
  • Or https://math.stackexchange.com/questions/3297618/calculating-lim-x-y-to0-0-fracx2yx2y4, https://math.stackexchange.com/questions/3257954/does-the-limit-lim-x-y-to0-0-fracx2yx2y4-exist if you want the same formula – Arnaud D. Oct 30 '20 at 22:17
  • @FreeZe you appear to be learning the methods that relate to limits at the origin in two variables. You asked earlier about when polar coordinates work, I left a comment there. If you have not already gone through it, I suggest becoming very, very comfortable with $\frac{x^2 y}{x^4 + y^2}$ As the answers here show, no change of variables is necessary in the problem you typed. – Will Jagy Oct 30 '20 at 22:30

1 Answers1

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HINT

By $x^2=u$ and $y=v$ we obtain

$$\left|\frac{xy^{2}}{x^{4}+y^{2}}\right|= \frac{\sqrt u v^{2}}{u^{2}+v^{2}} $$

user
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