$(\varepsilon, \delta)$ for continuity of a multivariable function

Question

Define $f : \Bbb R^2 \to \Bbb R^2$, $$f(x,y) = \begin{cases} 0 & (x,y) = 0 \\ \frac{xy(x^2-y^2)}{x^2+y^2} & (x,y) \ne 0 \end{cases} $$ Determine if $f$ is continuous.

At $\Bbb R^2_{\ne 0}$ we have that $f$ is continuous. In order to see if it's continuous at the origin I was approaching it using epsilon-delta. We have that $|f(x,y) - f(0,0)| < \varepsilon$ whenever $\sqrt{(x-0)^2+(y-0)^2} = \sqrt{x^2+y^2} < \delta$.

Now $$|f(x,y)-f(0,0)| = \frac{|xy||x^2-y^2|}{|x^2+y^2|} \leqslant \frac{|xy||x^2+y^2|}{|x^2+y^2|} =|xy|.$$

However I'm losing $x^2+y^2$ here since that's what I would need for $\delta$. What am I missing in my approach here? Is there another way I could bound this?

Answer 1

HINT: Remember that $|x|,|y|\le\sqrt{x^2+y^2}$.

Answer 2

one way of doing it would be to evaluate: $$L=\lim_{(x,y)\to(0,0)}\frac{xy(x^2-y^2)}{x^2+y^2}$$ now if we use polar coordinates we know that: $$x^2+y^2=r^2,x=r\cos\theta,y=r\sin\theta$$ s0: $$L=\lim_{r\to0}\frac{r^4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)}{r^2}=\lim_{r\to0}r^2\frac{\sin(2\theta)\cos(2\theta)}{2}=\lim_{r\to0}r^2\frac{\sin(4\theta)}{4}=0$$ and since $\theta$ does not affect this limit this is true whichever direction it is approached from, meaning the function is continuous for all $x,y\in\mathbb{R}^2$

Answer 3

For all $(x,y)\in\mathbb{R}^2$ we have $$ \left\lbrace \begin{align*} &2|xy|\le x^2+y^2\\[4pt] &|x^2-y^2|\le x^2+y^2\\[4pt] \end{align*} \right. $$ hence for all $(x,y)\in\mathbb{R}^2{\setminus}\{(0,0)\}$ we have $$ 2|f(x,y)| = \frac {(2|xy|)(|x^2-y^2|)} {x^2+y^2} \le \frac {(x^2+y^2)(x^2+y^2)} {x^2+y^2} = x^2+y^2 $$ which approaches $0$ as $(x,y)$ approaches $(0,0)$.