Proof for continuity: $f (x) = \left\{ \begin{array}{ll} \frac{2xy}{x^2+y^2} & (x,y)\neq(0,0) \\ 0 & \, (x,y)=(0,0) \\ \end{array} \right. $

Question

I have to prove the continuity of the following function:

$f (x) = \left\{ \begin{array}{ll} \frac{2xy}{x^2+y^2} & (x,y)\neq(0,0) \\ 0 & \, (x,y)=(0,0) \\ \end{array} \right. $

Case 1: $(x,y)\neq(0,0)$ is obviously continuous

Case 2: $(x,y)=(0,0)$, use $\epsilon$-$\delta$-criteria

$|f(x,y)-f(0,0)|=|f(x,y)|=|\frac{2xy}{x^2+y^2}|=\frac{|2xy|}{x^2+y^2}\leq2|xy|$

Look at a $\delta$-environment of $(0,0)$

$|(x,y)-(0,0)|<\delta \Rightarrow x^2+y^2<\delta^2$

$|x|<\delta, |y|<\delta \Rightarrow |xy|<\delta^2$, from where we can see

$|f(x,y)|\leq 2|xy|<2\delta^2\leq\epsilon^2 \Rightarrow \delta < \frac{\epsilon}{\sqrt{2}}$

Therefore the function has to be continuous for all $(x,y)\in\mathbb{R}^2$

My question is: Is the proof right and what can I improve in the clarity of my proof?

Answer 1

The function is not continuous at $(0,0).$ Note that

$$\lim_{x\to 0} f(x,\lambda x)=\lim_{x\to 0}\dfrac{2\lambda x^2}{(1+\lambda^2)x^2}=\dfrac{2\lambda}{1+\lambda^2}.$$

You use the inequality $|f(x,y)|\le 2|xy|$ but note that:

$$\frac{2|xy|}{x^2+y^2}=|f(x,y)|\leq 2|xy|\iff x^2+y^2\ge 1.$$ Since you are working on a neighbourhood of $(0,0)$ such inequality is not valid.