Continuity of $\frac{2xy}{x^2+y^2}$ at $(0,0)$ Given a Heaviside function
$$f(x,y)=\begin{cases}\frac{2xy}{x^2+y^2}, &x^2+y^2 \neq 0\\0 ,&x^2+y^2=0 \end{cases}$$     
Letting $a$ and $b$ be fixed constants, show that for all values of $a$ and $b$, including $0$, the one variable functions $g(x)=f(x,b)$ and $h(y)=f(a,y)$ are both continuous on the entire real line. And how to determine whether the function is continuous at $(0,0)$.
What I did was: substitute $x=b$ and $y=a$ in the function $f(x,y)$ to get


*

*$f(x,b)=\frac{2xb}{x^2+b^2}$

*$f(a,y)=\frac{2ay}{a^2+y^2}$


But I am unsure how to proceed from here. How should I take the limit of the function $f(x,y)$? How do I take care of the $x^2+y^2 \neq 0$?
 A: To prove your (two-variable) function is continuous at $(0,0)$, you have to prove $f(x,y)\rightarrow f(0,0)$ for $(x,y)\rightarrow(0,0)$, along any path. However, to prove it's not continuous at $(0,0)$, you have just to find one path that won't work.
Let's try $y=ax$, with $a\neq0$ and $x\neq0$:
$$f(x,y)=\frac{2ax^2}{x^2+a^2x^2}=\frac{2a}{1+a^2}$$
This is a constant $\neq0$, so as $x\rightarrow0$, $f(x,ax)$ does not convege to $f(0,0)$. Thus your function is not continuous at $(0,0)$.

Regarding the one variable functions $g(x)=\frac{2xb}{x^2+b^2}$ and $h(x)=\frac{2ax}{a^2+x^2}$, it's easier. Here is only the $h$ case, the other one is symmetrical:


*

*if $a\neq0$, then the function is continuous as the denominator never vanishes.

*if $a=0$, then the function is constant ($=0$) so it's also continuous.

A: Think of $x=r\cos(\theta)$ and $y=r\sin(\theta)$, then
$$
\begin{align}
\frac{2xy}{x^2+y^2}
&=\frac{2r^2\sin(\theta)\cos(\theta)}{r^2}\\
&=\sin(2\theta)
\end{align}
$$
That means no matter how close to $(0,0)$ you get, you can get any value in $[-1,1]$.
