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Epsilon-delta proof

  • Thread starter pbialos
  • Start date
pbialos
I have to prove the following:
[tex]f:B_1 (0)\subset\mathbb{R}^2\rightarrow\mathbb{R}\text { defined as }f(x)=\frac {1}{1-||x||}[/tex] 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:
[tex]\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[/tex]

note: [tex]B_1 (0)[/tex] 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.
 
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Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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One way to do this is to write x in polar coordinates. If x= (r, &theta;), then [tex]f(x)= \frac{1}{1-r} [/tex]. 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.
 

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