# Limits with absolute values

1. Mar 13, 2010

### math2010

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

Let f(x, y) = |xy|.

I want to prove that f is not continuous at (0,0).

3. The attempt at a solution

To prove that f is not continuous at (0,0) I think I need to show that

$$\lim_{(x, y) \to (0, 0)}|xy| \neq 0$$

I'm a little confused about the |absolute value| signs and I don't know know what to do with them. Can anyone help please? (there's nothin about this in my book)

2. Mar 13, 2010

### tiny-tim

Hi math2010!
I think it is continuous at (0,0)

Try proving that.

3. Mar 14, 2010

### math2010

Well the whole problem says "Show that $$f \notin C^1$$ at (0,0)."

In my course "$$f \in C^1$$" means that "$$f_x$$ and $$f_y$$ exist and are continuous".

So, I know that f(x, y) = |xy| is differentiable at (0,0). I want to prove that it does not belong to $$C^1$$, which is why I tried to show it's not continous. I'm very confused here

4. Mar 14, 2010

### jbunniii

I think the notion of $C^1$ is defined only if you are talking about some open set. I don't think it makes any sense to say "$f \in C^1$ at the point $(0,0)$".

It's true that $f$ is both continuous and differentiable at $(0,0)$.

However, $f$ is NOT differentiable at $(x,0)$ for $x \neq 0$, nor at $(0,y)$ for $y \neq 0$. Therefore if $U$ is an open set containing the origin, then $f \not\in C^1(U)$.

5. Mar 14, 2010

### tiny-tim

ah! that's not what you said originally

So f is in C0 (the continuous functions), but not in C1 (the continuously differentiable functions)
As jbunniii says, you need to show that f is not continuously differentiable in a neighbourhood of (0,0).

First step … what are the derivatives of f ?

6. Mar 14, 2010

### math2010

Since the derivative of the absolute value function is the function $$\frac{x}{|x|}$$ (where the denominator isn't zero), I guess the derivatives of f(x,y)=|xy| are $$f_{x}(x,y)= f_y(x,y)= \frac{xy}{|xy|}$$.

So what does this tell us?

7. Mar 14, 2010

### tiny-tim

No.

Differentiate without the ||, and then adjust it.

8. Mar 14, 2010

### math2010

without || it is:

So $$f_x (x,y) = y$$

And I adjust it to $$\frac{y}{|xy|}$$.

And $$f_y (x,y) = x$$ so it is $$\frac{x}{|xy|}$$. Is this correct?

9. Mar 15, 2010

### tiny-tim

No, the magnitude will just be y, so you need to multiply by the sign of xy (or xy/|xy| or |xy|/xy).

10. Mar 16, 2010

### math2010

ah, you mean $$f_x (x,y)= \frac{y}{|y|}$$ and $$f_y(x,y)= \frac{x}{|x|}$$?

11. Mar 16, 2010

### tiny-tim

But they both have magnitude 1, and they don't involve the sign of xy.

12. Mar 16, 2010

### math2010

What do you mean?

13. Mar 17, 2010

### math2010

I'm sorry, I don't really understand what you mean by saying "they don't involve the sign of xy". Do you mean it will be $$f_x(x,y) = y$$?

14. Mar 17, 2010

### tiny-tim

If it wasn't for the ||, fx(x,y) would be y.

The || will change the sign, but not the magnitude.

15. Mar 17, 2010

### math2010

So how does this enable us to show $$f \notin C^1$$?

16. Mar 17, 2010

### tiny-tim

I think f is differentiable at (0,0), but not continuously differentiable.