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Homework Help: How do you know what delta to choose

  1. Sep 3, 2009 #1
    Find the delta for the given epsilon. lim (1/x) =1 epsilon=.07

    2. Relevant equations

    3. The attempt at a solution
    I got to here .07526 >x-1> -.06542 so what one is me delta??
  2. jcsd
  3. Sep 3, 2009 #2
    I gotta run, but it will help whoever helps you if you say how you got there.
  4. Sep 3, 2009 #3


    Staff: Mentor

    You want a positive number for delta.
  5. Sep 3, 2009 #4
  6. Sep 3, 2009 #5
    Let [tex]\varepsilon > 0.[/tex] Assume the existence of a [tex]\delta > 0[/tex] such that [tex]0 < |x-1| < \delta.[/tex] Then

    [tex]\left|\frac{1}{x} - 1\right| = |x-1|\cdot\frac{1}{|x|} .[/tex]

    The |x-1| term already has a bound, so we need some bound on 1/|x|. Suppose that [tex]\delta \leq 1[/tex] so that
    [tex]|1| - |x| \leq |1-x| < \delta \Rightarrow |x| > 1 - \delta \Rightarrow \frac{1}{|x|} < \frac{1}{1-\delta}.[/tex]

    This implies that

    [tex]\left|\frac{1}{x} - 1\right| = |x-1|\cdot\frac{1}{|x|} < \frac{\delta}{1 - \delta}.[/tex]

    Requiring this last expression on the right to be [tex]\leq \epsilon,[/tex] we find that setting

    [tex]\delta = \min(1, \frac{\varepsilon}{\varepsilon + 1})\mbox{ ensures that } \left|\frac{1}{x} - 1\right| < \varepsilon.[/tex]

    Note: We chose delta to be less than or equal to 1 to make it easier to find a bound for 1/|x|, but we could have picked a number smaller than 1. Any number a > 1 would not work, since this would imply that |x-1| < a which would mean that x is in (1-a, 1+a) and since 1-a < 0 < 1 + a, division by zero could occur.

    Note: Plugging in epsilon = .07 gives the delta you were looking for.
    Last edited: Sep 3, 2009
  7. Sep 4, 2009 #6
    Oh, I see thank you snipez.
  8. Sep 4, 2009 #7
    but can you explain your steps a little more please, ive never seen it done like that before. up at the top how did X-1<delta turn into l1l-lxl<=l1-xl<=delta
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