How to prove that a limit of a two-variable function does not exist?

  • Thread starter Mike s
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I need to prove this limit does NOT exist. I know that the denominator cancels out, however my lecturer says the limit does not exist because we cannot find a circle with radius [itex]\delta[/itex] that contains the point (0,0), as the function is not defined for all y=x. Here is a little drawing of the domain.

Is my lecturer's explanation valid?


Thanks in advance,
Michael
 

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  • #3
I guess his explanation makes sense, but that's a funny way of putting it.
 
  • #4
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I guess his explanation makes sense, but that's a funny way of putting it.

Can you think of another way? If you do, please post
 
  • #5
SammyS
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I need to prove this limit does NOT exist. I know that the denominator cancels out, however my lecturer says the limit does not exist because we cannot find a circle with radius [itex]\delta[/itex] that contains the point (0,0), as the function is not defined for all y=x. Here is a little drawing of the domain.

Is my lecturer's explanation valid?

Thanks in advance,
Michael
Well, that limit certainly does exist --- at least it exists using the mufti-variable definition of limit with which I am familiar.

[itex]\displaystyle \text{If }x\ne y\,,\text{ then }\frac{x^2-y^2}{x-y}=\frac{(x-y)(x+y)}{x-y}=x+y\,.[/itex] No matter what continuous path you take (in the domain of the given function) to approach (0,0), x+y → 0 .

On the other hand, the function is not continuous at (x,y) = (0,0) because it's not defined at (x,y) = (0,0) .
 
  • #6
LCKurtz
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I think it is common for functions of two variables to be required to be defined in a punctured disk around (a,b) in order to claim a limit as ##(x,y)\to (a,b)## of the function exist. That is presumably required in this case, given the lecturer's statement.
 
  • #7
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Well, that limit certainly does exist --- at least it exists using the mufti-variable definition of limit with which I am familiar.

[itex]\displaystyle \text{If }x\ne y\,,\text{ then }\frac{x^2-y^2}{x-y}=\frac{(x-y)(x+y)}{x-y}=x+y\,.[/itex] No matter what continuous path you take (in the domain of the given function) to approach (0,0), x+y → 0 .

On the other hand, the function is not continuous at (x,y) = (0,0) because it's not defined at (x,y) = (0,0) .

Well, as far as I know, the limit does not have to be defined at the point itself (0,0), but there must exist at least one disc that contains it.
 
  • #8
LCKurtz
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Well, as far as I know, the limit does not have to be defined at the point itself (0,0), but there must exist at least one disc that contains it.

Yes, there must be a punctured disk around (a,b) in the domain of the function, which you don't have.
 
  • #9
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Yes, there must be a punctured disk around (a,b) in the domain of the function, which you don't have.

Alright, thanks a lot!
 
  • #10
HallsofIvy
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Well, that limit certainly does exist --- at least it exists using the mufti-variable definition of limit with which I am familiar.

[itex]\displaystyle \text{If }x\ne y\,,\text{ then }\frac{x^2-y^2}{x-y}=\frac{(x-y)(x+y)}{x-y}=x+y\,.[/itex] No matter what continuous path you take (in the domain of the given function) to approach (0,0), x+y → 0 .
Even the path y= x??

On the other hand, the function is not continuous at (x,y) = (0,0) because it's not defined at (x,y) = (0,0) .
 
  • #11
SammyS
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Even the path y= x??
y=x is not in the domain of [itex]\displaystyle \frac{x^2-y^2}{x-y}\ .[/itex]

It appears that Calculus textbooks don't agree on the definition of the limit for functions of more than one variable.
 

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