Homework Help: Continuity of multivariable functions question

1. May 30, 2010

QuanticEnigma

1. The problem statement, all variables and given/known data
Is the function f(x,y) defined by
f(x,y) = (yx^3 - 3y^3)/(x^2 + y^2), (x,y)!=(0,0)
=0, (x,y)=(0,0)
2. Relevant equations

3. The attempt at a solution
I changed f(x,y) into polar coordinates and found the limit as r->0, which is 0. I also found that f(x,0) = 0, f(0,y) = -3y^2 -> 0 as y->0, and f(x,x)= -infinity. However I'm not sure if I've done the question right or if this is enough working.

2. May 30, 2010

QuanticEnigma

Someone help!!!

3. May 30, 2010

tiny-tim

Welcome to PF!

Hi QuanticEnigma! Welcome to PF!

(have a ≠ and a ∞ and a θ and try using the X2 tag just above the Reply box )
Using polar coordinates is good, but it's not enough to check the limit only along the two axes.

Hint: use the fact that cosθ and sinθ are always ≤ 1.

4. May 30, 2010

QuanticEnigma

Ok, I've decided to cheat and use Mathematica to plot the function, and I can see that the function is continuous everywhere :tongue:
I think there must be some problem in my working, as the limits as f tends to (0,0) along the x and y axes is not the same as the limit along the line y=x...I'm confused

5. May 30, 2010

tiny-tim

uhh? isn't f(x,x) = (x2 - 3x)/2 ?

anyway, try what I said about θ.

6. May 30, 2010

QuanticEnigma

Yeah it is, my bad (hope I don't make that mistake in the exam :tongue:)

I don't really know what you mean by theta, I've already worked out the limit of f (in polar coordinates) as r tends to 0 to be zero, and I thought this implied that any line approaching (0,0) would have limit zero...I'm not sure though

Oh and by the way, every time I input latex code it always outputs f(x,y), even when I've copied and pasted it from other threads, any idea on how to fix this?

7. May 30, 2010

tiny-tim

I'm dubious about what you mean by "as r tends to zero".

I haven't seen your exact proof, but I assume you have an f(r,θ), and you can't prove continuity by only letting r -> 0 any more than you could by only letting x -> 0 and y -> 0 (separately).
It's a problem which has affected Preview ever since the last server update … you can fix it by clicking "Refresh" or "Reload" on your browser menu (do a tag search for "preview" for details).

Or just use the X2 tag.

8. May 30, 2010

QuanticEnigma

Ok, I have
$$f(x,y)=\frac{r^3\cos^3(\theta)r\sin(\theta)-3r^3\sin^3(\theta)}{r^2\cos^2(\theta)+r^2\sin^2(\theta)} = \frac{r^3(r\cos^3(\theta)\sin(\theta)-3\sin^3(\theta)}{r^2}=r(r\cos^3(\theta)\sin(\theta)-3\sin^3(\theta))$$

Since I have an r out the front of everything, i didn't worry about theta, because when r=0, f(r,$\theta$) would equal zero.

9. May 30, 2010

tiny-tim

(what happened to that θ i gave you? )

But that doesn't work if the inside of the bracket can be "infinitely" large, so you still have to prove that it can't be.

10. May 30, 2010

QuanticEnigma

Sorry, it's supposed to be $$f(r,\theta)$$ not $$f(x,y)$$

11. May 30, 2010

QuanticEnigma

Would sin$$\theta$$ and cos$$\theta$$ being $$\leq$$ 1 mean that the expression inside the bracket is between -1 and 1?

12. May 30, 2010

tiny-tim

Yup!

(well, between ±(r±3) )

13. May 30, 2010

QuanticEnigma

Sweet, thanks for the help