How to find/prove this two variable limit

[ozone]

It has been a while since I have done limis now, and I want to make sure I am doing this correctly.

I have the problem listed below
$(x-y)/(x+y) as (x,y) -> (0,0)$

In my mind I can gather that it will asymptote as you approach from either (-1,1) or (1,-1)

I also know that it will asymptote in opposite directions, and as far as I know this is what defines a function to have "no limit".

I am just wondering what I need to state algebraicly in order to "prove" this limit.

 

[DonAntonio]

It has been a while since I have done limis now, and I want to make sure I am doing this correctly.

I have the problem listed below
$(x-y)/(x+y) as (x,y) -> (0,0)$

In my mind I can gather that it will asymptote as you approach from either (-1,1) or (1,-1)

I also know that it will asymptote in opposite directions, and as far as I know this is what defines a function to have "no limit".

I am just wondering what I need to state algebraicly in order to "prove" this limit.

Choose $\,y=x\,$ and the limit is zero, but now chose $\,y=2x\,$ and the limit then is $\,\displaystyle{-\frac{1}{3}}$

Ergo: the limit doesn't exist.

DonAntonio

 

[haruspex]

It doesn't mean anything to let two variables tend to a limit simultaneously unless you specify the relationship between them. This is different from there being no limit. It has many possible limits and the question is incompletely specified.

 

[micromass]

It doesn't mean anything to let two variables tend to a limit simultaneously unless you specify the relationship between them. This is different from there being no limit. It has many possible limits and the question is incompletely specified.

Of course you can have limits of functions in two variables without having a relationship between them. This is basic multivariable calculus.

 

[haruspex]

Of course you can have limits of functions in two variables without having a relationship between them. This is basic multivariable calculus.

No. You can define the limit of the function as one variable tends to some value, and the limit of that as the other variable tends to some value. That's taking the limits to be in a particular order. If you swap the order you might or might not get the same result. Each is perfectly well defined in itself, but if you don't specify the order, and don't specify a relationship between them for approaching the limits simultaneously, it is simply not defined.

 

[micromass]

No. You can define the limit of the function as one variable tends to some value, and the limit of that as the other variable tends to some value. That's taking the limits to be in a particular order. If you swap the order you might or might not get the same result. Each is perfectly well defined in itself, but if you don't specify the order, and don't specify a relationship between them for approaching the limits simultaneously, it is simply not defined.

Did you take multivariable calculus??

For $f:\mathbb{R}^2\rightarrow \mathbb{R}$, we define $\lim_{(x,y)\rightarrow (a,b)} f(x,y)=L$ if

$$\forall \varepsilon >0: \exists \delta > 0: \forall (x,y)\in \mathbb{R}^2:~0<\|(x,y)-(a,b)\|_2<\delta~\Rightarrow~|f(x,y)-L|<\varepsilon$$

where we of course define $\|(x,y)\|_2=\sqrt{x^2+y^2}$.

 

[DonAntonio]

No. You can define the limit of the function as one variable tends to some value, and the limit of that as the other variable tends to some value. That's taking the limits to be in a particular order. If you swap the order you might or might not get the same result. Each is perfectly well defined in itself, but if you don't specify the order, and don't specify a relationship between them for approaching the limits simultaneously, it is simply not defined.

Either you didn't study several variables calculus or you already forgot: if the limit $\,\displaystyle{\lim_{(x,y)\to (x_0,y_0)}f(x,y)}\,$ exists then by definition

it must exist and be the same no matter how the variables approach the point $\,(x_0,y_0)\,$, and this is why

my first post proves the limit wanted in the OP doesn't exist.

DonAntonio

 

[haruspex]

I must have forgotten - sorry.