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Epsilon-delta proof

  1. Aug 8, 2005 #1
    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.
    Last edited by a moderator: Jan 7, 2014
  2. jcsd
  3. Aug 9, 2005 #2


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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).
  4. Aug 9, 2005 #3

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

    Regards, Paul.
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