TITLE: Continuity of $\frac{2xy}{x^2+y^2}$ at $(0,0)$
QUESTION [8 upvotes]: 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 $x^2+y^2 \neq 0$?

REPLY [3 votes]: 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 $[-1,1]$.