Is $f(x,y)$ continuous? I want to find out if this function is continuous:
$$(x,y)\mapsto \begin{cases}\frac{y\sin(x)}{(x-\pi)^2+y^2}&\text{for $(x,y)\not = (\pi, 0)$}\\0&\text{for $(x,y)=(\pi,0)$}\end{cases}$$
My first idea is that
$$\lim_{(x,y)\to(\pi,0)} |f(x,y)-f(\pi,0)|=\lim_{(x,y)\to(\pi,0)}\left|\frac{y\sin(x)}{(x-\pi)^2+y^2}\right|\\\le \lim_{(x,y)\to(\pi,0)}|y\sin(x)|\cdot \left|\frac{1}{(x-\pi)^2+y^2}\right| $$
Where the first term = 0 ($y=0$,$ \sin(\pi)=0$) but im not sure if this is leading me where I want.
Btw, I assume the function is continuous.

Let $u=x-\pi$. Then you're wondering about $$\lim_{(u,y)\to(0,0)}\frac{y\sin(\pi+u)}{u^2+y^2}$$ 
$$\lim_{(u,y)\to(0,0)}\frac{-y\sin u}{u^2+y^2}=$$
$$\lim_{(u,y)\to(0,0)}\frac{-yu}{u^2+y^2}\frac{\sin u}{u}$$
Can you show it doesn't go to zero? Look at $y=u$ and $y=-u$, for example.
