Suppose that ;

 

1)[math] f: R^2\rightarrow R[/math] such that : [math]f(x,y) =xy[/math]

 

2)The Euclidian norm of a vector [math] v=(u_{1}, u_{2})[/math] is defined as :

 

[math] ||v||_{Eu} =\sqrt{ u_{1}^2 +u_{2}^2}[/math]

 

3) The maxnorm of a vector [math] v=(u_{1},u_{2})[/math] is defined as :

 

[math] ||v||_{max} = max( |u_{1}|,|u_{2}|)[/math]

 

Where [math] u_{1},u_{2}[/math] belong to the real Nos R

 

Then prove :

 

[math] \lim_{(x,y)\to(1,1)} f(x,y) = 1[/math] ,with respect to both norms

What Euclidean Norm and Maxnorm have to do here? You can easily get the limit by substituting with x=1 and y=1, you will get lim f(x,y)=1*1=1.

What Euclidean Norm and Maxnorm have to do here? You can easily get the limit by substituting with x=1 and y=1, you will get lim f(x,y)=1*1=1.

 

 

That is not a proof is there??

O.K...... I was talking like that because f(x,y) is obviously continuous,

and lim f(x,y), as (x,y) tends to (xo,yo), is f(xo,yo).

If you need an exact proof, then find the left limit which will be f(xo - epsilon , yo - epsilon). And, find the right limit which will be f(xo + epsilon , yo + epsilon). Then, since both are equal then the limit is found.

O.K...... I was talking like that because f(x,y) is obviously continuous,

and lim f(x,y), as (x,y) tends to (xo,yo), is f(xo,yo).

If you need an exact proof, then find the left limit which will be f(xo - epsilon , yo - epsilon). And, find the right limit which will be f(xo + epsilon , yo + epsilon). Then, since both are equal then the limit is found.

 

 

How can you prove that left and right limits are equal??

 

How do you know that f(x,y) is continuous at ([math]x_{o},y_{o}[/math]),without proving it

1. You will have to check the Left Limit for the variable x. It will be lim f(x,y)=xo*y

2. Check the Right Limit for the variable x. It will be also lim f(x,y)=xo*y.

3. Therefore, the Left Limit of f(x,y), as x tends to xo, is equal to the Right Limit.

4. Repeat the previous, as y tends yo. And, you will get the same results.

5. Then lim f(x,y), as (x,y) tends to (xo,yo), is equal to f(xo,yo).

 

Note: There are some simple functions that can be considered continuous, unless it is required to prove so....... such as f(x)=x, f(x)=x^2.

1. You will have to check the Left Limit for the variable x. It will be lim f(x,y)=xo*y

2. Check the Right Limit for the variable x. It will be also lim f(x,y)=xo*y.

3. Therefore, the Left Limit of f(x,y), as x tends to xo, is equal to the Right Limit.

4. Repeat the previous, as y tends yo. And, you will get the same results.

5. Then lim f(x,y), as (x,y) tends to (xo,yo), is equal to f(xo,yo).

 

Note: There are some simple functions that can be considered continuous, unless it is required to prove so....... such as f(x)=x, f(x)=x^2.

 

You mean that:

 

IF [math]lim_{y\to y_{o}}(lim_{x\to x_{o}}f(x,y))=f(x_{o},y_{o})=x_{o}.y_{o}[/math]..................... THEN

 

[math] lim_{(x.y)\to(x_{o},y_{o})}f(x,y) =f(x_{o},y_{o})=x_{o}.y_{o}[/math]???

But, you shouldn't go the first step directly. You must use Left and Right limits (for both x and y).

But, you shouldn't go the first step directly. You must use Left and Right limits (for both x and y).

 

 

Now if this is not another pack of mathematical nonsenses what is it then?

 

You keep on writing non existing theorems and conclusions ,by simply imagining them.

Lose the attitude, triclino, we are a debate forum here and this is unacceptable.

 

If you have a claim to make, make it in a civil matter, or don't make it at all.