# Continuity of a two variable function

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.

## Answers and Replies

SammyS
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Change to polar coordinates.

SammyS
Staff Emeritus
Science Advisor
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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).

Understood, thanks.

SammyS
Staff Emeritus
Science Advisor
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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 .

Bacle2
Science Advisor
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 $\delta=\frac{1}{\sqrt{\epsilon}}$ [\QUOTE]

Should be $\frac{1}{(x^2+y^2)}>\frac{1}{\delta^2}$

since a>b implies $\frac{1}{a}< \frac{1}{b}$ when a,b ±0
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?[\QUOTE]

Limit as xn→x should equal f(x).

Sorry, I don't know how to correct quotes.

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