# The domain on which a function is continuous

1. Mar 31, 2013

### mrcleanhands

1. The problem statement, all variables and given/known data

Determine domain on which the following function is continuous
$$f(x,y)= \left\{\begin{array}{cc} \frac{x(x+1)y^{2}}{(x+1)^{2}+y^{2}} & (x,y)\neq(-1,0)\\ 1 & (x,y)=(-1,0) \end{array}\right.$$

2. Relevant equations

3. The attempt at a solution

Because the numerator component is a rational function it will be continuous for all $(x,y)\in\mathbb{R}$ except on (x,y)=(-1,0) and since the denominator has been set a domain of a single point (x,y)=(-1,0), it is also continuous in that domain. To test whether the whole piece-wise function is continous we should look for the limits as $(x,y)\rightarrow(-1,0)$.
$$\underset{(x,y)\rightarrow(-1,0)}{\lim}\frac{x(x+1)y^{2}}{(x+1)^{2}+y^{2}}$$ using the path y=mx
$$\underset{(x,y)\rightarrow(-1,-m)}{\lim}\frac{-1(-1+1)y^{2}}{m^{2}}=0$$

Approaching (-1,0) from the line y=mx gives us a limit of 0 which contradicts the limit of 1 for the 2nd component in the piece-wise function. Therefore the function as a whole is continous for all $(x,y)\in\mathbb{R}$ except on (x,y)=(-1,0).

I feel like something is not right with my reasoning or explanation. The function on the whole is continuous because it's defined for every x,y for all real numbers.

Last edited by a moderator: Apr 3, 2013
2. Mar 31, 2013

### Coelum

Your method is correct: the limit depends on how you approach the point (-1,0). A function defined for all (x,y) is not necessarily continuous everywhere - why should it be? Think of the following 1D example: f(x)=1/x forall x, f(0)=0.

3. Mar 31, 2013

### SammyS

Staff Emeritus
That all doesn't make sense.

The denominator of $\displaystyle \ \frac{x(x+1)y^{2}}{(x+1)^{2}+y^{2}}\$ is zero only at (x, y) = (-1, 0) . Therefore, we know that f(x,y) is continuous everywhere, except possibly at (x, y) = (-1, 0) .
The line y = mx, doesn't pass through (-1,0) , does it ?

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4. Apr 2, 2013

### mrcleanhands

yeah, you're right. so how can I find the limit for the second function then? There just isn't a limit?

5. Apr 2, 2013

### SammyS

Staff Emeritus
What second function ?

6. Apr 3, 2013

### mrcleanhands

I think I realized what was wrong with it. If I can't approach (-1,0) from the line y=mx then I can't approach it from the x axis, y axis, or z axis right?

So if I want to find the limit of this function
$\underset{(x,y)\rightarrow(-1,0)}{\lim}\frac{x(x+1)y^{2}}{(x+1)^{2}+y^{2}}$

I cannot test a line approaching (-1,0) by letting y=x as I can when looking for the limit when we approach (0,0).

Does this mean then that I should change the function to find the limit along different lines?

E.g. substitute (x-1) into x and then find the limit as (x,y) --> (0,0) will be the equivalent.

So

$$\underset{(x,y)\rightarrow(-1,0)}{\lim}\frac{x(x+1)y^{2}}{(x+1)^{2}+y^{2}} =\underset{(x,y)\rightarrow(0,0)}{\lim}\frac{(x-1)xy^{2}}{x{}^{2}+y^{2}}$$
right?

7. Apr 3, 2013

### jbunniii

Yes, this is correct.

Hint: consider applying the inequality $|xy| \leq \frac{1}{2}(x^2 + y^2)$ to
$$\left| \frac{(x-1)xy^2}{x^2 + y^2} \right|$$

8. Apr 3, 2013

### mrcleanhands

How did you get that inequality?

Also what do you meany by applying it to $$\left| \frac{(x-1)xy^2}{x^2 + y^2} \right|$$

9. Apr 3, 2013

### HallsofIvy

For any x, y, $(x+ y)^2= x^2+ 2xy+ y^2\ge 0$ so $-2xy\le x^2+ y^2$ and $(x- y)^2= x^2- 2xy+ y^2\ge 0$ so $2xy\le x^2+ y^2$.

So
$$\left|\frac{(x- 1)xy^2}{x^2+ y^2}\right|=\left|(x- 1)y\frac{xy}{x^2+ y^2}\right|\le\frac{1}{2}\left|(x-1)y\right|$$

10. Apr 3, 2013

### mrcleanhands

And this is related to epsilon delta proofing right? (struggling with at at the moment so can't really connect this)...

Btw I'm not supposed to use epsilon delta proofing to test the limit in this question. Just supposed to approach it from different lines which I haven't succeeded in doing thus far.

11. Apr 3, 2013

### jbunniii

OK, going back to your original expression:
$$\frac{x(x+1)y^{2}}{(x+1)^{2}+y^{2}}$$
what limit do you get if you approach $(-1,0)$ along the line $y = 0$?

12. Apr 3, 2013

### bayan

Removed

Last edited: Apr 3, 2013
13. Apr 5, 2013

### mrcleanhands

$f(x,0)=0$

14. Apr 5, 2013

### jbunniii

So what can you conclude?

15. Apr 5, 2013

### mrcleanhands

I didn't think you could conclude anything from that apart from what it is, approaching (-1,0) along the y axis produces a limit of 0 but that doesn't mean it's true for the whole function etc

16. Apr 5, 2013

### jbunniii

If the function was continuous at (-1,0), then the limit would have to be the same, no matter what direction you approach from. And what would the limit have to equal?

17. Apr 5, 2013

### mrcleanhands

0, same as approaching (-1,0) from the line y=0

18. Apr 5, 2013

### jbunniii

No, look at how the function is defined at (-1,0). What would the limit as (x,y) approaches (-1,0) have to be in order for the function to be continuous?

19. Apr 5, 2013

### mrcleanhands

I'm not following :(
What do you mean what would the limit have to be for the function to be continuous?

For the function to be continuous the limit would have to be the same as (x,y) approaches (-1,0) from any directional line...

20. Apr 5, 2013

### jbunniii

What is the value of the function at (-1,0)? Doesn't the limit have to match that value if the function is continuous?