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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.

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sin(pi) = 1 ? Are you sure ? –  Kalissar May 3 '13 at 13:55
    
edited, thanks. –  wegsehen May 3 '13 at 13:56

1 Answer 1

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.

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I dont know what to do next. For $y=u$ its $\lim \frac{-1\sin()}{2u})$ where the divisor seems to be growing faster which would lead to 0, but that doesn't seem like a proof to me. –  wegsehen May 3 '13 at 14:39
    
Try again. Remember that if $u\to 0$ then $$\frac{\sin u}u\to 1$$ You should get $-1/2$ and then $1/2$. –  Pedro Tamaroff May 3 '13 at 14:40

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