# Continuity of function in R^2

1. Feb 20, 2008

### jjou

Let $$f:\mathbb{R}^2\rightarrow\mathbb{R}$$ be $$f(0,0)=0$$ and $$f(x,y)=\frac{x|y|}{\sqrt{x^2+y^2}}$$ for $$(x,y)\neq(0,0)$$. Is f continuous at (0,0)?

I tried showing it WAS NOT continuous by finding sequences that converge to 0 but whose image did not converge to 0. I tried sequences of the form (ct, t) where c was a constant and t went to 0 as well as sequences of the form (t^c, t). Simple forms such as (t^c, t^c) or (1/t, 1/t) did not work either.

Then I tried to show it WAS continuous by showing it was lipschitz, which turned into a horribly horribly long expansion without a clear inequality - so I'm pretty sure this isn't the correct method.

Is there a method I am overlooking?

(Also, am I allowed to ignore the absolute value in the numerator if I restrict (x,y) to the first and second quadrants of $$\mathbb{R}^2$$?)

2. Feb 20, 2008

### Dick

Brute force is always good. Take sqrt(x^2+y^2)=r. Then |x|<=r and |y|<=r. So the absolute value of the numerator is less than r^2. So |f(x,y)|<=r. Now let r->0.

3. Feb 20, 2008

### jjou

Thanks!

So, I define a sequence $$(a_n)$$ such that, for each n, $$a_n=(x_n,y_n)\in\D_{1/n}(0,0)=\{a\in\mathbb{R}^2|d(a,0)=1\n\}$$. Then as $$n\rightarrow\infty$$ we have $$a_n\rightarrow0$$. Then for any n, $$\sqrt{x_n^2+y_n^2}=1/n$$ which implies that $$|x_n|\leq1/n$$ and $$|y_n|\leq1/n$$.

Then we have that $$f(a_n)\rightarrow f(0)=0$$ iff $$|f(a_n)|\rightarrow0$$.

$$|f(a_n)|=\left|\frac{x_n|y_n|}{\sqrt{x_n^2+y_n^2}}\right|=\frac{|x_n||y_n|}{\sqrt{x_n^2+y_n^2}}=n*|x_n||y_n|\leq n*\frac{1}{n}*\frac{1}{n} = \frac{1}{n}$$

which completes the proof since $$\frac{1}{n}\rightarrow0$$. Thus the function f is continuous.

Thanks again! :)

4. Feb 20, 2008

### jjou

Actually... I have to show this works for arbitrary sequence $$a_n$$ converging to (0,0). So I should define the sequence $$r_n=d(a_n,0)=\sqrt{x_n^2+y_n^2}$$ and the rest is the same.