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Continuity, but is it limits or delta-epsilon or neighborhood?

  1. Feb 22, 2010 #1
    EDIT: My presentation of this was pretty bad so I'm trying again.


    f: {0}U{1/N} --> R
    Where N is a natural number

    Defined piecewise:
    f(x) = 1/(x^2-x)

    I'm scared of this problem. Obviously, the function blows up with asymptotes at x=0,1 so plugging the holes piecewise with f(0)=f(1)=1 doesn't help with continuity.

    Last semester we covered limit-based continuity and delta-epsilon continuity. However, every single problem went from the real line to the real line. I don't see how you can do continuity with any of the three methods because of the discrete domain. The function doesn't have domain points to generate any of the range points that I want to be less than epsilon close.

    The graph of the function is going to be scatter-shot across R2 where every point is floating on its own...I can only concluded that the function isn't continuous ANYWHERE. How do you tackle this one?

    Thanks in advance.
    Last edited: Feb 22, 2010
  2. jcsd
  3. Feb 23, 2010 #2
    The definition of continuity hasn't changed for this smaller domain space; you just need to pay closer attention to the exact words.

    Let [tex]A\subset\mathbb{R}[/tex]. Then [tex]f: A \to \mathbb{R}[/tex] is continuous at [tex]a \in A[/tex] if, for every [tex]\varepsilon > 0[/tex], there exists [tex]\delta > 0[/tex] such that, whenever [tex]x \in A[/tex] satisfies [tex]|x - a| < \delta[/tex], then [tex]|f(x) - f(a)| < \varepsilon[/tex].

    To give an example: the function [tex]\iota: \mathbb{Q} \to \mathbb{R}[/tex] defined by [tex]\iota(x) = x[/tex] is continuous at every point of its domain [tex]\mathbb{Q}[/tex].

    You cannot speak of a function being continuous or discontinuous at a point where it's not defined, so in your case, you only have to answer whether the points [tex]0[/tex] and [tex]1/n[/tex] are points of continuity.
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