# Proving continuity of two-variable function

1. Nov 26, 2005

### twoflower

Hello to everybody.

Yesterday we wrote a fake test, in which I encountered problem I haven't benn able to solve so far.

The problem is given this way:

Let

$$f(x,y) := \left\{\begin{array}{cc}\frac{xy^3}{x^2+y^4} - 2x + 3y,&[x,y] \neq [0,0]\\0, &[x,y] = [0,0]\end{array}\right.$$

Find out, whether the function $f[/tex] in point [0,0] is continuous Well, to be continuous in origin, it has to have zero limit as [x,y] approaches [0,0]. I get $$\lim_{[x,y] \rightarrow [0,0]} \frac{xy^3}{x^2+y^4} - 2x + 3y$$ after converting to polar coordinates (which proved to be very useful when limiting two-variable functions) I have $$\lim_{r \rightarrow 0} \frac{r^4\cos \varphi \sin^3 \varphi}{r^2\cos^2 \varphi\ +\ r^4\sin^4 \varphi}\ -\ 2\lim_{r \rightarrow 0} r\cos \varphi\ +\ 3\lim_{r \rightarrow 0} r\sin \varphi = \lim_{r \rightarrow 0} r\cos\varphi \sin^3 \varphi\ \frac{r}{\cos^2 \varphi\ +\ r^2 \sin^4 \varphi}$$ The first part goes to zero so it would be sufficient to show that the second part is bounded. And that's where I got stuck. Horse sense tells me this: if [itex]\varphi$ is such that $\varphi$ is non-zero, the whole limit is zero. If $\cos \varphi$ is zero, then the limit is still zero...but I'm not sure of this and definitely it's not an acceptable mathematical solution. But I'm unable to prove this elegantly at this point...

Thank you very much for pointing me to the right direction.

Last edited: Nov 26, 2005
2. Nov 26, 2005

### math-chick_41

after you convert to polar and simplify all you have to do is find the limit as r goes to zero, phi can do anything. when r goes to zero your limit is zero.

3. Nov 26, 2005

### twoflower

Are you sure? Can it be so easily seen? I still have some doubts...

And one more question, can you see some further simplification there?