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Bounds of a function and limits

  1. Nov 8, 2007 #1
    1. The problem statement, all variables and given/known data
    A function is decreasing if f(x[tex]_{1}[/tex]) > f(x[tex]_{2}[/tex]) whenever x[tex]_{1}[/tex] < x[tex]_{2}[/tex], and x[tex]_{1}[/tex], x[tex]_{2}[/tex] [tex]\epsilon[/tex] [tex]\Re[/tex]

    a) Show that the set {f(x) : x < a} is bounded below
    b) Prove that lim (as x goes to a) f(x) = glb{f(x) : x < a}
    (hint: show that for any [tex]\epsilon[/tex] > 0, there exists [tex]\delta[/tex] > 0 such that f(a - [tex]\delta[/tex]) < c + [tex]\epsilon[/tex]

    i have no idea where to starttttt ><
    please helpp meeee
  2. jcsd
  3. Nov 8, 2007 #2


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    Staff Emeritus
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    You might START by stating the problem correctly. You give a definition of "decreasing function" but there doesn't appear to be a requirement that the "f" in the theorem be decreasing! Did you intend that f be decreasing? If so, make use of the definitions, since those are all you have! You know that x< a and you know that f is decreasing. "Bounded below" means "has a lower bound" which itself means that there is some number less than or equal to every number in the set. Can you make a guess at what a lower bound for {f(x)| x< a} must be? (Hint: look at f(a).)

    If a set of real numbers has a lower bound, then it must have a greatest lower bound. Use the definition of "greatest lower bound" in the hint.
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