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Simple absolute value problem with inequalities

  1. Aug 17, 2005 #1
    "Simple" absolute value problem with inequalities

    OK...Im totally stuck and could use some help :)
    given...for all e>0, d>0...the following holds
    |x-a|<d => |f(x) - f(a)| < e
    where f(x) = sqrt(x)

    how do I find d in terms of e?

    Thanks in advance
  2. jcsd
  3. Aug 17, 2005 #2


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    |x - a| = |[f(x) - f(a)][f(x) + f(a)| < e.|f(x) + f(a)| < d
    I don't think you can get simpler than that.
  4. Aug 17, 2005 #3


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    I am going to assume you meant: for all e > 0 there is some d > 0 such that |[itex]x-a[/itex]|< d implies |[itex]\sqrt x - \sqrt a[/itex]| < e.

    ([itex]\sqrt x - \sqrt a[/itex])2 < e2

    [itex]x + a - 2\sqrt{x a}[/itex] < e2

    [itex]x - a + 2(a-\sqrt{x a})[/itex] < e2

    [itex]x - a[/itex] < e2 - [itex]2(a-\sqrt{x a})[/itex]

    If [itex]x - a[/itex] > 0 then d = e2 - [itex]2(a-\sqrt{x a})[/itex]

    (So d depends on x, and I guess that's okay.)

    I need to think about the case where [itex]x - a[/itex] < 0.
  5. Aug 17, 2005 #4


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    It is simplest to note that:
    and proceed from there.
  6. Aug 17, 2005 #5
    well...I end up with something like [tex]|x-a|<d => |x-a|=e|\sqrt{x}+\sqrt{a}|[/tex]...And thats where I am stuck on :(

    yah...d has to be independent of x...its one of those proving limit typa thing. I am just allowed to assume a = 4 first
  7. Aug 17, 2005 #6


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    If you are attempting to prove that [itex]\sqrt{x}[/itex] is continuous for all positive values of x, then d does not have to be independent of d. That's only true for uniform continuity.

    If you have [itex]|x-a| |x-a|=e|\sqrt{x}+\sqrt{a}|[/itex] and x is "sufficiently close to a", say, |x-a|< 1/2, so that a- 1/2< x< a+ 1/2, what can you say about [itex]\sqrt{x}+ \sqrt{a}[/itex]?
  8. Aug 17, 2005 #7
    hehe...how did you get [tex]|x-a|<d => |x-a|=e|\sqrt{x}+\sqrt{a}|[/tex]???
    I think it should be [tex]|x-a|<d => |x-a|<e|\sqrt{x}+\sqrt{a}|[/tex]
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