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Another epsilon-delta proof

  1. Oct 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Similar to the problem I just posted, here's another one:

    Suppose [tex]A[/tex] is the set of all numbers that can be written as [tex]\dfrac{1}{n}[/tex], where [tex]n \in \mathbb{N}[/tex].

    Define a function [tex]f[/tex] such that

    [tex]
    \begin{equation*}
    f(x) = \left\{
    \begin{array}{cc}
    0 & : x \in A\\
    x & : x \not \in A
    \end{array}
    \end{equation*}
    [/tex]

    Prove that

    [tex]
    \lim_{x \to a} f(x) = a
    [/tex]


    2. Relevant equations



    3. The attempt at a solution

    Proof:

    For [tex]x \in (-\infty, 0)\cup(1,\infty) \Rightarrow f(x) = x[/tex]

    which is a continuous function. Thus

    [tex]
    \lim_{x \to a} f(x) = f(a) = a
    [/tex]


    Now consider the interval [tex](0, 1][/tex]. Suppose [tex]a[/tex] is in this interval. If [tex]a \not \in A[/tex], then [tex]\delta[/tex] can be chosen sufficiently small so that the open interval [tex](a-\delta, a+\delta)[/tex] does not contain any [tex]x \in A[/tex]. In this subinterval, [tex]f(x) = x[/tex] and

    [tex]
    \lim_{x \to a} f(x) = f(a) = a
    [/tex]

    Now suppose [tex]a \in A[/tex]. Then [tex]a[/tex] can be written as [tex]a = \dfrac{1}{n}[/tex].
    Choose [tex]\delta = |\dfrac{1}{n}-\dfrac{1}{n+1}| = |\dfrac{1}{n(n+1)}|[/tex].
    Thus [tex]\delta[/tex] is the minimum distance to the next [tex]x \in A[/tex].

    Once I get here, I'm stuck... I'm trying to show that if delta is the minimum distance to the next x in A, then for all [tex]x \not = a[/tex] in that subinterval, [tex]f(x) = x[/tex]. But I'm confused about what to do next. I'm not sure what this implies with regards to epsilon.

    I'm also not sure what to do for x = 0.
     
    Last edited: Oct 18, 2009
  2. jcsd
  3. Oct 18, 2009 #2
    I've had a think about it and I think this is what it means...

    If [tex]\epsilon \geq \dfrac{1}{n(n+1)}[/tex] then take [tex]\delta = \dfrac {1}{n(n+1)}[/tex].

    If [tex]0 < \epsilon < \dfrac{1}{n(n+1)}[/tex] then take [tex]\delta = \epsilon[/tex].

    Ugh, the way I've chosen to do it seems complicated...
     
  4. Oct 18, 2009 #3
    can anyone confirm that what i'm on the right track here?
     
  5. Oct 18, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    . There is nothing much else to be done. For that delta, as you say, [itex]f(x)= x[/itex] so |f(x)- a|= |x- a|. Now, if we call that "delta" [itex]\delta_1[/itex], we can take [itex]\delta[/itex] to be the smaller of [itex]\delta_1[/itex] and [itex]\epsilon[/itex]. If [itex]0< |x-a|< \delta[/itex], then, since |x- a|< delta so |f(x)- a|= |x- a|< [itex]\epsilon[/itex]

    If a= 0, then, for any [itex]\epsilon> 0[/itex], there exist x in A less than [itex]\epsilon[/itex] so that f(x)= 0. But now |f(x)- a|= |f(x)| so that's not a problem.
     
  6. Oct 18, 2009 #5
    Thanks very much for the clarification.
     
  7. Oct 18, 2009 #6
    Both of these problems are trivial in light of one lemma. The set up for the lemma is this: Let [tex]f[/tex] be a function on a set [tex]S[/tex] and suppose that [tex]\lim_{x \to a} f(x) = b[/tex]

    Now let [tex]

    \begin{equation*}
    f_{a}(x) = \left\{
    \begin{array}{cc}
    f(x) & : x \neq a\\
    \alpha & : x = a
    \end{array}
    \end{equation*}
    [/tex]

    Where [tex]\alpha[/tex] is arbitrary (in fact, we don't even need to define [tex]f_{a}[/tex] at [tex]a[/tex]). What can you say about [tex]\lim_{x \to a}f_{a}[/tex]?
     
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