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Epsilon-delta test for continuity

  1. Dec 2, 2008 #1
    Hi all!

    I´m having some trouble finding a delta for f(x)=(x-2)² using the epsilon-delta definition for fixed epsilon and x_0. Here´s what I come up with:


    now I divide by the second term and define my delta with the fixed epsilon and x_0, but there also appears this x which by definition should not be there, should it?


    so, how can I make x disappear and thereby get some delta for my epsilon?

    thanks a lot in advance!
  2. jcsd
  3. Dec 2, 2008 #2


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    Try the triangle inequality:
    [tex]|x + x_0 - 4| = |x - (x_0 + 4)| \le \cdots [/tex]
  4. Dec 2, 2008 #3
    Try the triangle inequality:
    LaTeX Code: |x + x_0 - 4| ???=??? |x - (x_0 + 4)| \\le \\cdots

    Did you mean:

    [tex]|x + x_0 - 4| = |x + (x_0 - 4)|<=|x|+|x_0-4|[/tex]

    but this does not mean that:
    [tex]|x-x_0||x + (x_0 - 4)|<=|x-x_0||x|+|x-x_0||x_0-4|[/tex]

    but we now do not know if |x-x_0||x|+|x-x_0||x_0-4| is less than epsilon, dow we?

    So how does the triangle inequality apply here?
  5. Dec 2, 2008 #4


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    Hmm, you are right, I got my signs mixed up. OK, so maybe a better attempt would be to write
    [itex]|x + x_0 - 4| = |(x - x_0) + (2x_0 - 4)| \le |x - x_0| \cdot 2|x_0 - 2| \le 2 \delta |x_0 - 2|[/itex]
    Just play around to get (x - x0) and some constants.
  6. Dec 2, 2008 #5
    hmmm, I still cannot follow what you´ve done:


    but why should it also be less or equal |(x-x_0)|.2|(x_0-2)|

    and what about the term |x-x_0| in |x-x_0||x+x_0-4|< "epsilon" or we just multiply by it in the end?
  7. Dec 2, 2008 #6
    I don't mean to hijack this thread, but why can't you have an x in your delta?
  8. Dec 2, 2008 #7
    I think the deltas and the epsilons should only be dependent on the point in which you examine the function for continuity, i.e. from x_0

    but it´s not so easy, at least to me :)
  9. Dec 2, 2008 #8
    [tex]|(x-2)^2 - (x_0 - 2)^2| = |x^2 - x_0^2 -4x + 4x_0| = |x^2 - x_0^2 -4(x - x_0)| =
    |(x-x_0)(x+x_0) -4(x - x_0)| = |x-x_0| |x+x_0 - 4| [/tex]

    Now, the point x_0 must be contained in some closed interval [a,b], and in this interval obviously every x in this interval (excluding b) is less than b. So we now have:

    [tex]|x-x_0| |x+x_0 - 4| < |x-x_0| |2|b| - 4| < \delta |2|b| - 4| = \epsilon[/tex]

    So we can take [tex] \delta = \frac{\epsilon}{|2|b| - 4|}[/tex]
  10. Dec 2, 2008 #9
    Your delta will depend on x0. A function f is continuous if for all x0 in the domain of f and for all ε > 0 there exists a δ > 0 such that for all x in the domain of f, |x - x0| < δ implies that |f(x) - f(x0)| < ε. The order of the quantifiers here is important; you can pick your δ to depend on both x0 and ε.

    There is another condition called uniform continuity, where for all ε > 0 there exists a δ > 0 such that for all x0 and x in the domain of f, |x - x0| < δ implies that |f(x) - f(x0)| < ε. The order of the quantifiers is crucial here: f(x) = (x - 2)2 is continuous but not uniformly continuous.
  11. Dec 2, 2008 #10
    If you restrict the x to a neighborhood of x_0 lying completely in a closed interval then it doesn't have to depend on [tex]x_0[/tex]
    Last edited: Dec 2, 2008
  12. Dec 4, 2008 #11
    Is this thread solved? It doesn't look like it to me. I don't know about JG89's argument. Can you just do that, restrict x-x0, by some constant and call it a day? Doesn't delta need to be chosen so that it will hold for any x? I think JG89's delta can be broken if you choose an x "far out" enough.

    I was working on a proof for pretty much the same thing, except f(x)=x2. My T.A. solved it by breaking the function into cases. When x0>0, =0, <0, then one can choose delta<x0 (for x0>0) and delta<-x0 (for x0<0) which implies

    x+x0 < 2x0, for the positive case the negative is almost identical

    and can go one to solve

    |x2-x02| = |x-x0||x+x0| <

    (delta)|x-x0| < delta|2x0| which we have constructed to be less than any epsilon. Solve for delta and find
    delta = epsilon/2x0

    As I write this I wonder, is it right? Is JG89's definition sufficient?
  13. Dec 4, 2008 #12
    My calculus book says it's okay to choose delta to depend on end points of an interval (because then if it's continuous in that interval, it is automatically uniformly continuous) as long as you restrict the x to [tex]a\le x \le b[/tex].

    As for a function [tex] f(x) = x^2 [/tex], my textbook says this, in case you're interested:

    "The function [tex]f(x) = x^2[/tex] for an infinite x-interval is definitely not uniformly continuous. It is clear that small changes in x can produce arbitrarily large changes in [tex]x^2[/tex] if only x is large enough. A glance at a table of squares of integers x shows how successive squares are spaced further and further apart as x increases. If, however, we only consider pairs of values [tex]x[/tex] and [tex]x_0[/tex] belonging to a fixed finite closed interval [tex][a,b][/tex], we can find a uniform modulus of continuity. Indeed, for [tex]|x-x_0| < \delta [/tex] we have:

    [tex]|f(x) - f(x_0)| = |x^2 - x_0^2| = |x-x_0||x+x_0| \le 2|x-x_0|(|b| + |a|) < 2\delta(|b| + |a|) = \epsilon [/tex]

    if we take [tex] \delta = \frac{\epsilon}{2(|b|+|a|)}[/tex]"
  14. Dec 4, 2008 #13

    |x-a||x+a-4|=< |x-a|(|x|+|a|+4) ,since by using the triangle inequality twice we have

    |x+a-4|=< (|x|+|a|+4)....................................................................................1

    Now consider values of x near a,so take |x-a|<1 , a very common trick when considering ε-δ proofs for polynomials.

    But |x-a|<1 ===> |x|-|a|<1 , since |x|- |a |=<|x-a|, and

    (|x|+|a|+4)=< 2|a|+5.

    Now we want |x-a||x+a-4|<ε hence |x-a||x+a-4|=< |x-a|(2|a|+5)<ε if

    |x-a|< ε/ (2|a|+5)

    So now given ε>0 take δ=min{ 1, ε/(2 |a|+5)} ===> δ=<1 and δ=< ε/ (2|a|+5)

    AND if |x-a|<δ ====> { |x-a|<1 and |x-a|<ε/ (2|a|+5) } which will lead us to:

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