The domain on which a function is continuous

[mrcleanhands]

Homework Statement

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

Homework Equations

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 continuous 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 continuous 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.

 

[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.

 

[SammyS]

Homework Statement

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

Homework Equations

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.

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) .

To test whether the whole piece-wise function is continuous 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 continuous 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.

The line y = mx, doesn't pass through (-1,0) , does it ?


Use the following for LaTeX renderings.

For inline LaTeX use:
[itex] [/itex]    or    ## ##

For stand alone, \displaystyle use:
[tex] [/tex]    or     $$ $$

 

[mrcleanhands]

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

 

[SammyS]

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

What second function ?

 

[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?

 

[jbunniii]

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?

Yes, this is correct.

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

 

[mrcleanhands]

Yes, this is correct.

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

How did you get that inequality?

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

 

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

 

[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.

 

[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 $y = 0$?

 

[bayan]

Removed

 

[mrcleanhands]

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 $y = 0$?

$f(x,0)=0$

 

[jbunniii]

$f(x,0)=0$

So what can you conclude?

 

[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

 

[jbunniii]

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

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?

 

[mrcleanhands]

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?

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

 

[jbunniii]

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

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?

 

[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...

 

[jbunniii]

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...

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

 

[mrcleanhands]

Ok I see that the limit would have to match the function. It's undefined at (-1,0).

But that does not mean that the limit does not exist right? since the function may get infinitely close to some point but not touch it.

 

[SammyS]

Ok I see that the limit would have to match the function. It's undefined at (-1,0).

But that does not mean that the limit does not exist right? since the function may get infinitely close to some point but not touch it.

The piecewise defined function, f(x,y), which appears in the OP is defined everywhere, even at (-1,0). In fact, f(-1,0) = 1 .

The expression, $\ \displaystyle \frac{x(x+1)y^{2}}{(x+1)^{2}+y^{2}}\$ is defined for all x & y, except at the point, (-1,0). And the limit of this expression as (x,y) → (-1,0) does exist -- but you have not proved that. This limit needs to be equal to f(-1,0) , if f is to be continuous at (-1,0) .

 

[mrcleanhands]

Okay I follow that. I get one of the limits as (x,mx) --> 0 as 0 so that means the function is not continuous.

However, it's weird because the question asks me to show that the limit exists or doesn't and support my claim without epsilon delta proofing. I keep getting 0 when approaching it from different lines but how do I ever know if there isn't some weird line which when approaching (0,0) doesn't = 0. I don't see how I can answer this question without epsilon delta unless I find one line which doesn't approach 0 - showing the function has no limit...

 

[SammyS]

Okay I follow that. I get one of the limits as (x,mx) --> 0 as 0 so that means the function is not continuous.

However, it's weird because the question asks me to show that the limit exists or doesn't and support my claim without epsilon delta proofing. I keep getting 0 when approaching it from different lines but how do I ever know if there isn't some weird line which when approaching (0,0) doesn't = 0. I don't see how I can answer this question without epsilon delta unless I find one line which doesn't approach 0 - showing the function has no limit...

The following link discusses finding the same limit for this very same expression.

https://www.physicsforums.com/showthread.php?t=683042

See if you can follow it. The basic idea was to shift the function one unit to the right,
then change to polar coordinates and find the limit as r → 0 .

 

[mrcleanhands]

I did ... dammit and it took me a few minutes to answer the question on 3 lines where as I spent something like 30min+ figuring it out before with tons of lines and arbitrary stabs.

 

[mrcleanhands]

Yes, this is correct.

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

I'm still curious why this was brought up however...

 

[SammyS]

Yes, this is correct.

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

I'm still curious why this was brought up however...

It's one way to attack the problem, although it does appear most suitable for an ε-δ proof . Notice that x² + y² = r² .

In general, to find a limit as (x,y) → (0,0), using polar coordinates is very useful. You then have a limit in one variable: r → 0 .

 

[jbunniii]

I'm still curious why this was brought up however...

Yes, this is correct.

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

This allows you to easily see that
$$\lim_{(x,y) \to (0,0)} \frac{(x-1)xy^{2}}{x^{2}+y^{2}} = 0$$
To see this, we note that
$$\begin{align}
\left| \frac{(x-1)xy^{2}}{x^{2}+y^{2}} \right| &=
\left| \frac{(x-1)y}{x^2 + y^2} \right| |xy| \\
&\leq \left| \frac{(x-1)y}{x^2 + y^2} \right| \left|\frac{1}{2}(x^2 + y^2)\right| \\
&= \frac{1}{2} \left| (x-1)y\right| \\
&\leq \frac{1}{2} |x-1| |y|
\end{align}$$
and the final expression clearly approaches $0$ as $(x,y) \to (0,0)$.

 

[mrcleanhands]

Ok, got it. Thank you!

This forum is awesome.

:D