Continuous function under 2 variables

[gipc]

Hello,
Is the following function continuous?
http://img192.imageshack.us/img192/7566/captureywn.jpg

If so, how do I prove it?

 

[micromass]

Maybe you can start by writing

\frac{3x^4-2y^4}{x^2-y^2}=\frac{x^4}{x^2-y^2}+2\frac{x^4-y^4}{x^2-y^2}

 

[gipc]

I can't see how that gets me any closer to prove the function is continuous (but maybe that's because I'm stupid)

 

[micromass]

Well, I claim that the left term of the sum is always continuous (I see a difference of two squares in the numerator). So it suffices to prove that the right is continuous...

 

[gipc]

maybe I neglected to say that the question is about the point (0,0), and I'm not sure how to prove it mathematically. had it been a single variable function i would have just taken the limit as x->0, but I don't know which steps to take when talking about f(x,y). So I'm basically helpless :(

 

[gipc]

Maybe you can start by writing

\frac{3x^4-2y^4}{x^2-y^2}=\frac{x^4}{x^2-y^2}+2\frac{x^4-y^4}{x^2-y^2}

sorry for the bump

How can I see that this is Continuous? Should I move to polar coardinates?

I need to "formally" submit this assignment and I can't quite put it into words.
The condition that |x|=|y| is giving me much trouble. Even if I take rsina, rcosa it doesn't look promising because of the absolute value condition. I can't just say that, well, we have r^2*abs(sin^4 x/(sin^2 x - cos^2 x)) and the absolute value is bounded and r goes to zero so the limit is zero- I have to take into consideration that |x|=|y| and I'm a little lost.

I'll appreciate more help.

 

[micromass]

Sorry for not answering before, I must have missed your first post.

The only reason to write

<br /> \frac{3x^4-2y^4}{x^2-y^2}=\frac{x^4}{x^2-y^2}+2\frac{x^4-y^4}{x^2-y^2}<br />

is because it makes it easier to check whether our function is continuous. Since it reduces the question to seeing if \frac{x^4}{x^2-y^2} is continuous.

But actually, this function is not continuous. For example, it is not continuous is (1,1). To check this, you need to find a sequence which will converge to (1,1), but whose image does not converge to (1,1). Finding such a sequence is not that difficult...

 

[gipc]

The function is not continuous?
That came as a surprise to me.

What do you mean to find a sequence which will converge but whose image doesn't?
Is it like taking to different "routes" for example, define y=x^2 and then the limit becomes
x^4/(x^2-x^4) as x->0?

 

[Tedjn]

Yes, it is like taking different routes and finding different limits for the function. If the function is continuous, the function will converge to its actual value along every route.

 

[gipc]

I've tried some ordinary routes like y=kx^2 or x=y but couldn't show that the limit at (0,0) is different than 0.

Can someone please give further help?

 

[micromass]

Try to take a route to (1,1) (try to take a route parallel to the y-axis), what does that give you?

 

[gipc]

Why should I look at (1,1)? We need to check the point (0,0), isn't it?

 

[micromass]

I though the question was, "Is the function continuous", well, I say the function isn't continuous at (1,1)...
It probably isn't continuous at (0,0) to, but that's a bit harder...

 

[gipc]

any ideas :(

 

[micromass]

Take a route to (1,1) (hint: parallel to the y-axis) and prove that f is not continuous at (1,1)...

 

[gipc]

but i need to prove it is not continuous at (0,0), the other points are irrelevant.

or maybe it is continuous ?

 

[micromass]

I really don't see why the other points are irrelevant.

I think I have a prove that the function is not continuous at (0,0).
Take \epsilon=1 and take \delta arbitrary. In the ball B((0,0),\delta), we can always find a point of the form (x,x). And if we take a route parallel to the x-axis through (x,x), then the image of this route will go through infinity. Thus there is no way we can keep the image of B((0,0),\delta) under 1...