Continuity of function in R^2

jjou

Let [tex]f:\mathbb{R}^2\rightarrow\mathbb{R}[/tex] be [tex]f(0,0)=0[/tex] and [tex]f(x,y)=\frac{x|y|}{\sqrt{x^2+y^2}}[/tex] for [tex](x,y)\neq(0,0)[/tex]. 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 [tex]\mathbb{R}^2[/tex]?)

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.

jjou

Thanks!

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

Then we have that [tex]f(a_n)\rightarrow f(0)=0[/tex] iff [tex]|f(a_n)|\rightarrow0[/tex].

[tex]|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}[/tex]

which completes the proof since [tex]\frac{1}{n}\rightarrow0[/tex]. Thus the function f is continuous.

Thanks again! :)

jjou

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

8-)