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 $\displaystyle (x,y)-->(0,0)$
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 $\displaystyle (x,y)-->(0,0)$
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 $\displaystyle ~2|ab|\le a^2+b^2$.
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.