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

[Mike s]

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 $\delta$ 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

 

[Ravi Mandavi]

Rationalize it

 

[NewtonianAlch]

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

 

[Mike s]

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

 

[SammyS]

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 $\delta$ 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.

$\displaystyle \text{If }x\ne y\,,\text{ then }\frac{x^2-y^2}{x-y}=\frac{(x-y)(x+y)}{x-y}=x+y\,.$ 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) .

 

[LCKurtz]

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.

 

[Mike s]

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

$\displaystyle \text{If }x\ne y\,,\text{ then }\frac{x^2-y^2}{x-y}=\frac{(x-y)(x+y)}{x-y}=x+y\,.$ 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.

 

[LCKurtz]

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.

 

[Mike s]

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!

 

[HallsofIvy]

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

$\displaystyle \text{If }x\ne y\,,\text{ then }\frac{x^2-y^2}{x-y}=\frac{(x-y)(x+y)}{x-y}=x+y\,.$ 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) .

 

[SammyS]

Even the path y= x??

y=x is not in the domain of $\displaystyle \frac{x^2-y^2}{x-y}\ .$

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