# Continuity of a two variable function

1. Oct 12, 2011

### eptheta

I have a function in x and y, and I was trying to figure out if it was continuous or not.
$f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}$

As far as I know, the only problematic point in the domain is (x,y)=(0,0) so I tried to use the $\epsilon,\delta$ definition.
My proposed limit at (0,0) being 0,
$\frac{|x^2-y^2|}{(x^2+y^2)^2} < \epsilon$
when $\sqrt{x^2+y^2}<\delta$

The epsilon term $< \frac{|x^2|+|y^2|}{(x^2+y^2)^2} =\frac{x^2+y^2}{(x^2+y^2)^2} =\frac{1}{(x^2+y^2)} = \frac{1}{\delta^2}$
therefore, take $\delta=\frac{1}{\sqrt{\epsilon}}$ and the definition is fulfilled.

1.Wolfram alpha tells me the limit doesn't exist, the plot clearly shows me the graph tends towards infinity at (0,0), so there's something wrong with my concepts.
Could someone point out what ?
2.Is the fact that the limit exists at every point a sufficient condition for continuity?

Thanks.

2. Oct 12, 2011

### SammyS

Staff Emeritus
Change to polar coordinates.

3. Oct 12, 2011

### SammyS

Staff Emeritus
Suppose you let (x,y) approach (0,0) along the x-axis, so that y=0.

f(x,0) = x2/x4 = 1/x2.

limx→0(1/x2) = +∞ .

By the Way, f(x,y) can't be continuous at (0,0), if it's not defined at (0,0).

4. Oct 12, 2011

### eptheta

Understood, thanks.

5. Oct 13, 2011

### SammyS

Staff Emeritus
A more general way to look at this is with polar coordinates.

x = r cos(θ), y = r sin(θ) .

x2 + y2 = r2

$\displaystyle \frac{x^2-y^2}{(x^2+y^2)^2}=\frac{r^2(\cos^2(\theta)-sin^2(\theta))}{(r^2)^2}=\frac{\cos(2\theta)}{r^2}$

When y = ±x, cos(2θ) = 0. So approaching (0,0) along a line of slope ±1, gives a result of zero.

Approaching (0,0) along any other line gives a result of +∞.

Try approaching (0,0) along the curves: y = x ± 2x3 .

6. Oct 13, 2011

### Bacle2

Last edited: Oct 13, 2011