# Limits with absolute values

• 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)

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. 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 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 $f \not\in C^1(U)$.

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 ? 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? 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. 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? 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).

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 $$f_y(x,y)= \frac{x}{|x|}$$?

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

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

What do you mean?

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

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

The || will change the sign, but not the magnitude. So how does this enable us to show $$f \notin C^1$$?

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