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Uniform Continuity of 1/x^2 on various sets

  1. Nov 21, 2008 #1
    1. The problem statement, all variables and given/known data

    Show that f(x)=[tex]\frac{1}{x^{2}}[/tex] is uniformly continuous on the set [1,[tex]\infty[/tex]) but not on the set (0,1].

    2. Relevant equations



    3. The attempt at a solution

    I've been working at this for at least 2 hours now, possibly 3, and I can't say I really have much of any idea about it. I've primarily been looking at the fact that a function f: A[tex]\rightarrow[/tex] [tex]\Re[/tex] is not uniformly continuous iff [tex]\exists[/tex] some [tex]\epsilon_{0}[/tex] > 0 and sequences xn and yn where |xn-yn| [tex]\rightarrow[/tex] 0 but |f(xn)-f(yn)|[tex]\geq[/tex] [tex]\epsilon[/tex]0.

    My thought was to somehow say that, for part one, because all sequences must be [tex]\geq[/tex] 1, they must converge to the same thing and therefore |f(xn)-f(yn)|= 0 for some n, but I'm not sure if I can for various reasons. In fact, I can't even really come up with any sequences in (0,1] to show that the function is not uniformly continuous on that set. I can think of another way to do that one, but I really want to see some sequences to do it to prove that it works in that case.

    Any help is greatly appreciated.
     
  2. jcsd
  3. Nov 21, 2008 #2

    Mark44

    Staff: Mentor

    Look at the graph of f(x) = 1/x^2. For the interval x >= 1, the graph is fairly level, so if |x1 - x2| is small, |f(x1) = f(x2)| is small as well. OTOH, the graph is very steep for x close to zero, so |x1 - x2| being small doesn't guarantee that |f(x1) - f(x2)| will also be small.
     
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