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Show that f(x) is unbounded

  1. Mar 30, 2015 #1
    1. The problem statement, all variables and given/known data

    Show that the function
    [tex]
    f(x) =\\

    \begin{cases}
    \frac{1}{x} &\quad 0 < x \leq 1\\
    0 &\quad x = 0
    \end{cases}
    [/tex]

    is unbounded.
    2. Relevant equations
    If f is bounded, |f(x)| <= M for all x in f's domain.
    3. The attempt at a solution

    I tried arguing by contradiction: suppose there is such an M. Then |f(x)| = f(x) <= M. But if f(x) < 1/M, f(x) > M. But I get stuck because that means this particular choice of bound does not work. Instead, choose N > M. But then f(x) < 1/N makes f(x) > N. There might be a bound, and I'm having trouble proving that there is not a bound.
     
    Last edited by a moderator: Mar 30, 2015
  2. jcsd
  3. Mar 30, 2015 #2

    Mark44

    Staff: Mentor

    If x < 1/M, then f(x) > M.
    You can do this directly, without resorting to a proof by contradiction.
     
  4. Mar 30, 2015 #3
    But then can't you choose N > M and still be bounded? I'm guessing that I have to show that we can always take x small enough so that there's no M that will always satisfy the condition f(x) <= M for x in (0, 1]. But I don't know how to do that.
     
  5. Mar 30, 2015 #4

    Mark44

    Staff: Mentor

    I think you might be confusing this problem with one in which ##\lim_{x \to \infty} f(x) = \infty##.
    Yes. Let some "large number" M be given. Then if 0 < x < 1/M, then f(x) > M. That's all you need to say.
     
  6. Mar 30, 2015 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    @bigplanet401: The statement that ##f(x)## is bounded means that there is a number ##M## such that for all ##x##, ##|f(x)|\le M##. I think it would help your thought process if you wrote a careful denial of that. The statement that ##f(x)## is unbounded means..., then prove that.
     
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