# Continuous function

1. Aug 7, 2008

### asi123

1. The problem statement, all variables and given/known data

Hey.
How can I show that this is a Continuous function?

2. Relevant equations

3. The attempt at a solution

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2. Aug 7, 2008

### HallsofIvy

Staff Emeritus
First, it is clearly continuous everywhere except possibly (0,0). I presume you have no trouble with that. In order to show that it is also continuous at (0,0), use the definition of continuous: that $\lim_{(x,y)\rightarrow (0,0)} f(x,y)= f(0,0)$. Now, what does that vertical line in the formula mean? You appear to have
$$\frac{x^2 y^2}{x^2y^2|(x-y)^2}$$
I don't understand that "|".

3. Aug 7, 2008

### asi123

Where do u see a vertical line? it's a plus sign
I know that I need to show that $\lim_{(x,y)\rightarrow (0,0)} f(x,y)= f(0,0)$, any idea about how to show it?

4. Aug 7, 2008

### morphism

Have you tried simplifying the expression for f(x,y) when (x,y)$\neq$(0,0)?

5. Aug 7, 2008

### asi123

What do you mean by simplify?

6. Aug 8, 2008

### HallsofIvy

Staff Emeritus
Okay, so you need to show that the limit, as (x,y) goes to (0,0) of
$$\frac{x^2y^2}{x^2y^2+ (x- y)^2}$$
is 1, the value of the function at (0,0).

There doesn't seem to me to be any good way to simplify that (multiplying out, reducing the fraction, etc. just like you did in algebra)

The best way to find the limit of a function of two variables, going to (0,0) is to change to polar coordinates. That way a single variable, r, measures the distance from (0,0). In polar coordinates, $x= rcos(\theta)$ and y= $rsin(\theta)$ so $x^2y^2= r^4 sin^2(\theta)cos^2(\theta)$ and $(x- y)^2= (rcos(\theta)- rsin(\theta))^2= r^2(cos(\theta)- sin(\theta))^2$. The function, in polar coordinates, is
$$\frac{r^4 sin^2(\theta)cos^2(\theta)}{r^2(sin^2(r^2theta)cos^2(\theta)sin^2(\theta)+ (cos(\theta)-sin(\theta))^2}= r^2\left[\frac{sin^2(\theta)cos^2(\theta)}{r^2sin^2(\theta)cos^2(\theta)+ (cos(\theta)- sin(\theta)^2}\right]$$
and now the point is that because of that "r2" factored out, as r goes to 0, the whole thing goes to 0 no matter what $\theta$ is.

Now, that does cause a serious problem as far as showing that this function is continuous is concerned!

7. Aug 8, 2008

Got ya, 10x.