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I need to show $$\lim_{(x, y) \rightarrow (0,0)} \frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}} = 0 $$ Not really sure how to go about this without using the epsilon delta definition, which I would prefer not to. Any sort of help is appreciated.

Edit: I do have the inequality:

$$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}} \le \frac{xy(x^2+y^2)}{(x^2+y^2)^{3/2}} = \frac{xy}{(x^2+y^2)} $$

Robert Z
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Okeh
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3 Answers3

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Let $(x,y) \not =(0,0)$.

$0\le |\dfrac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}| \le \dfrac{|xy||x^2+y^2|}{(x^2+y^2)^{3/2}} $

$\le \dfrac{(x^2+y^2)^2}{(x^2+y^2)^{3/2}}= (x^2+y^2)^{1/2}.$

Used:

1)$|x^2-y^2| \le |x^2+y^2|;$

2)$x^2+y^2 \ge |xy|$.

Peter Szilas
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Use polar coordinates: by letting $x=r\cos(\theta)$ and $y=r\sin(\theta)$, we have that $$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}= \frac{r^4\sin(2\theta)\cos(2\theta)}{2r^3}=\frac{r\sin(4\theta)}{4}.$$ Therefore $$\left|\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}\right|\leq \frac{r}{4}$$ and the right-hand side goes to $0$ as $r\to 0$.

P.S. Your equality is too "weak" for the job: by taking $y=x$ the right-hand side is $\frac{1}{2}$.

Robert Z
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  • Thanks, are using polar coordinates quite a common approach to solving such limits? Haven't encountered their use before – Okeh Sep 22 '19 at 11:01
  • Yes, it is. You can find more examples in this site. For example: https://math.stackexchange.com/questions/3297618/calculating-lim-x-y-to0-0-fracx2yx2y4 – Robert Z Sep 22 '19 at 11:08
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Let $y=tx$. Then

$$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}=\frac{x^2tx^2(1-t^2)}{x^3(1+t^2)^{3/2}}=x\frac{t(1-t^2)}{(1+t^2)^{3/2}}.$$

As the fraction remains bounded for all values of $t$, the limit is $0$.

  • I think this is not true, because the existsence and equality of the limits along all of the lines through the origin won't imply the existence of the original limit. (See, for example, my old question: https://math.stackexchange.com/questions/2741516/is-there-an-mathbbr2-to-mathbbr-function-with-the-following-limit-pr ) – Botond Sep 22 '19 at 12:21
  • @Botond: I never spoke about a limit along all lines in the sense of constant $t$. $(x,y)$ and $(x,t)$ are equivalent 2D representations, as $t$ spans all of $\mathbb R$. Feel free to retract your downvote. –  Sep 22 '19 at 12:38
  • I see. I thought that you fixed $t$. But I did not downvote your answer, so I can't retract it, just give you an upvote :) (I prefer to not upvote and ask for clarification over downvote, especially in the case of smarter people's answers) – Botond Sep 22 '19 at 12:45
  • @Botond: sorry for suspecting you :-) –  Sep 22 '19 at 12:50