Continuity

**Ulysses** · September 13th 2010, 12:47 PM

I must say if the function is continuous in the point (0,0). Which is $\displaystyle\lim_{(x,y) \to{(0,0)}}{f(x,y)}=f(0,0)$

The function:

$f(x,y)=\begin{Bmatrix} (x+y)^2\sin(\displaystyle\frac{\pi}{x^2+y^2}) & \mbox{ if }& y\neq{-x}\\1 & \mbox{if}& y=-x\end{matrix}$

I think its not continuous at any point, cause for any point I would ever have a disk of discontinuous points, but I must prove it. And I wanted to do so using limits, which I think is the only way to do it.

$\displaystyle\lim_{(x,y) \to{(0,0)}}{(x+y)^2\sin(\displaystyle\frac{\pi}{x^ 2+y^2})}$

What should I do? should I use trajectories? the limit seems to exist, as the sin oscilates between -1 and 1, and the other part tends to zero.

**tonio** · September 13th 2010, 02:58 PM

Originally Posted by Ulysses

I must say if the function is continuous in the point (0,0). Which is $\displaystyle\lim_{(x,y) \to{(0,0)}}{f(x,y)}=f(0,0)$

The function:

$f(x,y)=\begin{Bmatrix} (x+y)^2\sin(\displaystyle\frac{\pi}{x^2+y^2}) & \mbox{ if }& y\neq{-x}\\1 & \mbox{if}& y=-x\end{matrix}$

I think its not continuous at any point, cause for any point I would ever have a disk of discontinuous points, but I must prove it. And I wanted to do so using limits, which I think is the only way to do it.

$\displaystyle\lim_{(x,y) \to{(0,0)}}{(x+y)^2\sin(\displaystyle\frac{\pi}{x^ 2+y^2})}$

What should I do? should I use trajectories? the limit seems to exist, as the sin oscilates between -1 and 1, and the other part tends to zero.

$\lim\limits_{(x,y)\to (0,0)} (x+y)^2=0\,\,\,and\,\,\,\sin\left(\frac{\pi}{x^2+y ^2}\right)$ is bounded, so the limit does exists and equals..., and thus the function isn't continuous at the origen.

Tonio

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