# Continuity of two-variable function

1. Oct 16, 2005

### twoflower

Hi, I have some troubles understanding the basic facts about investigating the continuity of two-variable functions.
Our professor gave us very simple example to show us the basic facts:

Very important is that projections are continuous, it means

$$\pi_1 :[x,y] \longmapsto x$$

$$\pi_2 :[x,y] \longmapsto y$$

are continuous.
Now let's have this function:

$$f(x,y) = \sqrt{x^2+y^2}$$

It's continuous on the whole $\mathbb{R}^2$. Why?

1. Step:
Projections are continuous

2. Step:

$$x,y \longmapsto x^2 + y^2$$...continuous

3. Step:
Square root is continuous on $[0,\infty)$
What I don't fully understand is the second step. I can't see why, from the fact that projections are continuous, we can say that

$$x^2 + y^2$$

is continous. Of course I didn't expect any other result, when you look at it it's obvious that it will be continuous, but I just can't see the rigorous mathematical background.

You know, I see it is some equivalent of the limit of product of single variable function (? is it the right expression ?), but I don't think it's sufficient. I would need some equivalent for two-variable functions...

Hope you understand my problem :)

Thank you very much.

Last edited: Oct 16, 2005
2. Oct 18, 2005

### vanesch

Staff Emeritus
I'll give it a try, but I'm not a mathematician, so I might get it wrong ; feel free to correct me if that's the case.
Consider the function $$a \rightarrow a^2$$. That's a polynomial and hence continuous.
Now, consider the first function:
$$(x,y) \rightarrow x \rightarrow x^2$$
This is a projection, followed by our squaring, so that's a continuous function from (x,y) to x^2.
Next, consider the second function:
$$(x,y) \rightarrow y \rightarrow y^2$$
This is also a projection, followed by squaring, so that's a continuous function too, from (x,y) to y^2.
The sum of two continuous functions, is continuous, so if we add our two functions, we have $$(x,y) \rightarrow x^2 + y^2$$
should be continous.
Did I sneak in something ?

3. Dec 14, 2005

### siddharth

Twoflower,

Let the point $M_0 (x_0,y_0)$ belong to the domain of the function $f(x,y)$. Then the function $z=f(x,y)$ is continous at the point $M_0 (x_0,y_0)$ if we have

$$\lim_{\substack{x\rightarrow x_0\\y\rightarrow y_0}} f(x,y) = f(x_0,y_0)$$

and (x,y) approaches $M_0 (x_0,y_0)$ in arbitary fashion all the while remaining in the domain of the function.

Now if $x=x_0 + \Delta x , y = y_0 + \Delta y$ then the above condition can be rewritten as

$$\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} f(x_0 + \Delta x,y_0 + \Delta y) = f(x_0,y_0)$$

$$\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} [f(x_0 + \Delta x,y_0 + \Delta y) - f(x_0,y_0)] = 0$$

or
$$\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} \Delta z = 0$$

So for $$z=\sqrt{x^2+y^2}$$,
what is
$$\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} \Delta z$$ for any point in the domain?

Last edited: Dec 14, 2005
4. Dec 14, 2005

### JasonRox

Doesn't get any easier than this.

It's very simliar, if not the same, as continuity for a single variable.

5. Dec 15, 2005

### HallsofIvy

Staff Emeritus
Actually, it does- put it in polar coordinates and it IS continuity for a single variable!

The point made earlier, by vanesch, is that you cannot be certain of getting the correct limit of
$$\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} \Delta z$$

By simply taking $\Delta y$ going to zero and then taking $\Delta x$ going to 0. However, you can find the limit by seeing what happens when $(\Delta x, \Delta y)$ is very close to (0,0). In polar coordinates that is measured by the single variable r. What do you get if you put this problem in polar coordinates?