# 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