Let $\displaystyle f(x,y) = xy$($\displaystyle \dfrac{x^2-y^2}{x^2+y^2}$). show that $\displaystyle |f(x,y)|\le\dfrac{1}{2}(x^2+y^2)$, and hence prove that $\displaystyle f(x,y)$approaches a limit as $\displaystyle (x,y)-->(0,0)$

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- Sep 28th 2010, 02:31 PMwopashuiprove f approaching to a limit
Let $\displaystyle f(x,y) = xy$($\displaystyle \dfrac{x^2-y^2}{x^2+y^2}$). show that $\displaystyle |f(x,y)|\le\dfrac{1}{2}(x^2+y^2)$, and hence prove that $\displaystyle f(x,y)$approaches a limit as $\displaystyle (x,y)-->(0,0)$

- Sep 28th 2010, 02:39 PMPlato
Notice that $\displaystyle ~|2xy|\le x^2+y^2$.

- Sep 29th 2010, 08:41 AMRunty
I'm not exactly sure how we'd be able to use that, Plato. This particular question is tricky, and our textbook doesn't provide very good examples (or rather ANY examples) on how to do this.

If anyone could provide some more details, that'd be great. - Sep 29th 2010, 10:27 AMPlato
- Sep 29th 2010, 12:49 PMRunty
Wow, I am slow. Then again, proof-work has never been my forte in mathematics (I'm more of a number cruncher).

One thing, though: wouldn't it be $\displaystyle ~2|xy|\le x^2+y^2$ instead of $\displaystyle ~|2xy|\le x^2+y^2$?

I say this because of the inequality $\displaystyle ~2|ab|\le a^2+b^2$. - Sep 29th 2010, 12:58 PMPlato
- Sep 30th 2010, 11:08 AMRunty
I found another potential answer to the question through searching online. This answer uses polar coordinates, though I'm not sure how accurate it is.

Let $\displaystyle x=r\cos t$ and $\displaystyle y=r\sin t$.

$\displaystyle \left| {xy\dfrac{{x^2 - y^2 }}

{{x^2 + y^2 }}} \right|=\dfrac{r^4|\sin t\cos t(\cos^2t-\sin^2t)|}{r^2}=\dfrac{r^2|\sin2t\cos2t|}{2}<\dfra c{r^2}{2}<\epsilon$ if

$\displaystyle r<\sqrt{2\epsilon}$ is independent of the angle $\displaystyle t$, so the limit is 0.

The person who posted this didn't give all the steps I could've wanted, so expect it to have some errors. - Oct 2nd 2010, 07:28 PMwopashui