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Homework Help: Epsilons and Deltas Book Error? Or My Error?

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Given f(x) = mx + b, m > 0, L = (m/2) + b, xo = 1/2 , [itex]\epsilon = c >0[/itex] find (a) an open interval on which the inequality
    |f(x) - L| < [itex]\epsilon[/itex]
    holds. Then find (b) [itex]\delta[/itex] such that 0 < |x - xo| < [itex]\delta\Rightarrow[/itex] |f(x) - L| < [itex]\epsilon[/itex]

    Here is my problem with the book's solution. Since the condition [itex]\epsilon=c>0[/itex] was given, I only used the right-hand-side of the inequality:

    [itex]-c<|f(x)-L|<c[/itex] because to me it did not make sense to solve the inequality under a condition that cannot be. Instead, I chose to write the above inequality as:


    But the text gave answer of (a) [itex](\frac{1}{2}-\frac{c}{m}, \frac{c}{m}+\frac{1}{2})[/itex] and (b) [itex]\delta = c/m[/itex]

    Why did they use the left-hand-side of the inequality if it was given that c > 0 ?
  2. jcsd
  3. Sep 13, 2009 #2
    The book is right, but I don't know why it introduces the unneeded term c. I'm not sure if they are trying to trick you, but epsilon, or c for that matter, is just an arbitrary positive number. Look at the inequality |f(x) - L| < e again (e is just an arbtirary positive number; we usually think of it as very small). You interpreted this correctly in the previous limit question I helped you out with. All this inequality is saying is that f(x) is within a distance e from L.

    Remember, |f(x) - L| < e is equivalent to L - e < f(x) < L + e. If you choose to write it in the latter form, you have to drop the absolute value signs (which I think is what tripped you up).
  4. Sep 13, 2009 #3
    Oh yeah! We define epsilon to be a positive number.
    That wAs stupid of me!

  5. Sep 13, 2009 #4
    what course is this
  6. Sep 13, 2009 #5
    This from a Calculus Textbook. Self Study.
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