Proving Non-Continuity at (0,0) for f(x,y) = |xy|

[math2010]

Homework Statement

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

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

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)

 

[tiny-tim]

Hi math2010!

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

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

I think it is continuous at (0,0)

Try proving that. ​

 

[math2010]

Hi math2010!


I think it is continuous at (0,0)

Try proving that. ​

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

 

[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).

 

[tiny-tim]

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

ah! that's not what you said originally

So f is in C⁰ (the continuous functions), but not in C¹ (the continuously differentiable functions)

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

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 ? ​

 

[math2010]

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 ? ​

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?

 

[tiny-tim]

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|}.

No.

Differentiate without the ||, and then adjust it.

 

[math2010]

No.

Differentiate without the ||, and then adjust it.

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?

 

[tiny-tim]

without || it is:

So f_x (x,y) = y

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

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

 

[math2010]

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

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

 

[tiny-tim]

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

 

[math2010]

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

What do you mean?

 

[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?

 

[tiny-tim]

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

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

 

[math2010]

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

 

[tiny-tim]

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