# Thread: Functions of two variables continuity question

1. ## Functions of two variables continuity question

Consider the following functions each of which is defined on the x - y plane

f1(x) = (x-y)/(x+y) if x + y is not 0 and otherwise f1(x,y) = 0

f2(x,y) = (xy)/(x^2 + y^2) if (x,y) is not (0,0) and otherwise f2(0,0) = 0

f3(x,y) = (x^3 - y^3)/(x^2 + y^2) if (x,y) is not (0,0), and otherwise f3(0,0) is 0

Which of these is continuous

A) none B) f1 only C) f2 only D) f3 only E) all three

I know the defination of continuity for a single variable is the lim as x-> a of f(x) = f(a)

So i assume for two variables it should be the lim as (x,y) -> (a,b) of f(x,y) = f(a,b)

But I am not sure how to figure this out can someone help out please

2. Its pretty easy, in each function you have $\displaystyle f(0,0)=0$

Take the limit for each one as $\displaystyle (x,y) \to (0,0)$

If the limit = 0 , then the function is continuous at (0,0)

Do you have any problems in evaluating these limits?

3. But the lim (x,y) ->(0,0) is 0 for all them is it.. I mean Since the function is peicewise defined... If i plug in (0,0) then each one gives 0/0 Thats what i do not understand.

I understand it is in indeterminant form. But we haven't used polar coordinates. As far as algebra goes, nothing cancels when factored.

With the path rule should I approach (0,0) from the left and (0,0) from the right? I am not sure what that will do..

4. You should know that 0/0 is undefined
the limit still exist or does not exist
we do not know anything about it
you should use another methods to evaluate it
such as: polar coordinates, 2-path rule , algebric methods .. etc

5. Well since it is in indeterminant form in there a l'hopitals rule for a function of two variables? Because I am not sure how to do a path argument

6. No.
There is no L'Hospital's Rule for the multivariable limits.

For the first one, when y=0 , the limit = 1. But when y=x, the limit=0.
So the limit does not exist.
Hence, f1 is not continuous at (0,0) ..

7. I see I undersand. You found two paths in which the limits are not equal so the limit does not exist because if the limit did exist, we would get the same limit no matter what path we took.

So for f2(x,y) = xy/(x^2 + y^2)

The path y = x gives me the limit = 1/2

But the path y = 0 gives me the limit = 0

This means f2(x,y) is not continuous at (0,0) either

And for f3(x,y) = x^3 - y^3/(x^2 + y^2)

the path y = x gives me the limit = 0
the path y = 1 gives me the limit = -1

So f3(x,y) is not continuous at (0,0) either. So none of them are.

is this correct?

8. Correct.
But you should know that this method does not work if the limit exist.

As an example: $\displaystyle \lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2+y^2}}$

If you take any path, the limit always will be 0
Here, you should think about another way to deal with this limit
What I want to say is that do not think in every problem that the limit does not exist and try to search for paths that have different values !