Epsilon-Delta Proof for f(x)=1/(1-||x||) Continuity and Unboundedness

[pbialos]

I have to prove the following:
$$f:B_1 (0)\subset\mathbb{R}^2\rightarrow\mathbb{R}\text { defined as }f(x)=\frac {1}{1-||x||}$$ prove that f is continuous and not bounded.
I can see the geometrical explanation for this to happen, and i suppose i should prove the continuity using a delta-epsilon method, but i am stuck doing this.
The delta-epsilon proof would go like this:
$$\forall\epsilon>0,\exists\delta>0/ ||(x-x_0,y-y_0)||<\delta\rightarrow |{\frac {1}{1-||(x,y)||}-\frac {1}{1- ||(x_0,y_0)||}|<\epsilon$$

note: $$B_1 (0)$$ means the ball of radius 1 centered at the origin, that is to say, all the points (x,y) such that ||(x,y)||<1.
I know i am supposed to use this on the epsilon-delta proof, but even though i have no idea how to do it.

Any help i greatly appreciated.
Thanks you, Paul.

 

[HallsofIvy]

One way to do this is to write x in polar coordinates. If x= (r, &theta;), then $$f(x)= \frac{1}{1-r}$$. It should be easy to prove that that is continuous for r< 1 but not bounded (as r-> 1).

 

[pbialos]

Great

Great, now I think I understand how to prove it. Thank you very much for all your help.

Regards, Paul.