Limits with absolute values

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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

[tex]\lim_{(x, y) \to (0, 0)}|xy| \neq 0[/tex]

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)
 

Answers and Replies

  • #2
tiny-tim
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Hi math2010! :smile:
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) :confused:

Try proving that. :smile:
 
  • #3
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Hi math2010! :smile:


I think it is continuous at (0,0) :confused:

Try proving that. :smile:

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

In my course "[tex]f \in C^1[/tex]" means that "[tex]f_x[/tex] and [tex]f_y[/tex] 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 [tex]C^1[/tex], which is why I tried to show it's not continous. I'm very confused here :confused:
 
  • #4
jbunniii
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I think the notion of [itex]C^1[/itex] is defined only if you are talking about some open set. I don't think it makes any sense to say "[itex]f \in C^1[/itex] at the point [itex](0,0)[/itex]".

It's true that [itex]f[/itex] is both continuous and differentiable at [itex](0,0)[/itex].

However, [itex]f[/itex] is NOT differentiable at [itex](x,0)[/itex] for [itex]x \neq 0[/itex], nor at [itex](0,y)[/itex] for [itex]y \neq 0[/itex]. Therefore if [itex]U[/itex] is an open set containing the origin, then [itex]f \not\in C^1(U)[/itex].
 
  • #5
tiny-tim
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Well the whole problem says "Show that [tex]f \notin C^1[/tex] at (0,0)."
ah! that's not what you said originally :wink:

So f is in C0 (the continuous functions), but not in C1 (the continuously differentiable functions)
In my course "[tex]f \in C^1[/tex]" means that "[tex]f_x[/tex] and [tex]f_y[/tex] 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 [tex]C^1[/tex], which is why I tried to show it's not continous. I'm very confused here :confused:
As jbunniii :smile: 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 ? :smile:
 
  • #6
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ah! that's not what you said originally :wink:

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


As jbunniii :smile: 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 ? :smile:
Since the derivative of the absolute value function is the function [tex]\frac{x}{|x|}[/tex] (where the denominator isn't zero), I guess the derivatives of f(x,y)=|xy| are [tex]f_{x}(x,y)= f_y(x,y)= \frac{xy}{|xy|}[/tex].

So what does this tell us? :rolleyes:
 
  • #7
tiny-tim
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Since the derivative of the absolute value function is the function [tex]\frac{x}{|x|}[/tex] (where the denominator isn't zero), I guess the derivatives of f(x,y)=|xy| are [tex]f_{x}(x,y)= f_y(x,y)= \frac{xy}{|xy|}[/tex].
No.

Differentiate without the ||, and then adjust it. :wink:
 
  • #8
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No.

Differentiate without the ||, and then adjust it. :wink:
without || it is:

So [tex]f_x (x,y) = y[/tex]

And I adjust it to [tex]\frac{y}{|xy|}[/tex].

And [tex]f_y (x,y) = x[/tex] so it is [tex]\frac{x}{|xy|}[/tex]. Is this correct? :confused:
 
  • #9
tiny-tim
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without || it is:

So [tex]f_x (x,y) = y[/tex]

And I adjust it to [tex]\frac{y}{|xy|}[/tex].
No, the magnitude will just be y, so you need to multiply by the sign of xy (or xy/|xy| or |xy|/xy).
 
  • #10
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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 [tex]f_x (x,y)= \frac{y}{|y|}[/tex] and [tex]f_y(x,y)= \frac{x}{|x|}[/tex]?
 
  • #11
tiny-tim
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But they both have magnitude 1, and they don't involve the sign of xy.
 
  • #12
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But they both have magnitude 1, and they don't involve the sign of xy.
What do you mean?
 
  • #13
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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 [tex]f_x(x,y) = y[/tex]?
 
  • #14
tiny-tim
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If it wasn't for the ||, fx(x,y) would be y.

The || will change the sign, but not the magnitude. :smile:
 
  • #15
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So how does this enable us to show [tex]f \notin C^1[/tex]?
 
  • #16
tiny-tim
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I think f is differentiable at (0,0), but not continuously differentiable.
 

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