1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

More Epsilon Delta Limits

  1. Nov 3, 2009 #1
    Given: limit of (sin x)/x as x --> 0 and that ε = .01
    Problem: Find the greatest c such that δ between zero and c is good. Give an approximation to three decimal places.

    Equations:

    0 < |x - a| < δ
    0 < |f(x) - L| < ε


    Attempt:

    0 < |x - 0| < δ
    0 < | sin(x)/x - 1| < ε

    0 < | sin(x)/x - 1| < .01
    0 < | sin(x)/x| < 1.01
    0 < |sin(x)| < 1.01|x|
    0 < |sin(x)| < 1.01δ
    0 < |sin(x)|/1.01 < δ
    Since sin(x) is going to range between -1 and 1, the greatest value for δ is 1/1.01. But, this answer isn't correct. The correct answer is .245, and I don't know how to get that.
     
  2. jcsd
  3. Nov 3, 2009 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What do you mean by "δ between zero and c is good"?

    My guess is that you mean:

    "Find the greatest [itex]c[/itex] such that

    [tex]\left|\frac{\sin x}{x} - 1 \right| < \epsilon[/tex]

    whenever

    [tex]0 < |x - 0| < \delta \leq c[/tex]."

    Is that right?

    [It seems clearer to dispense with [itex]c[/itex] and simply ask "how big can [itex]\delta[/itex] be?"]

    Your problem is the following step:

    While the upper bound in the second line is true, it doesn't buy you anything. In fact, it's true no matter what [itex]\epsilon[/itex] is, because [tex]|\sin(x)/x|[/tex] never exceeds 1.

    Also, the lower bounds are wrong: both [tex]|\sin(x)/x - 1|[/tex] and [tex]|\sin(x)/x|[/tex] do equal 0 for some values of [itex]x[/itex]. We don't want/need to disallow that possibility.

    Try writing

    [tex]\left|\frac{\sin x}{x} - 1\right| < 0.01[/tex]

    in the following equivalent way

    [tex]-0.01 < \frac{\sin x}{x} - 1 < 0.01[/tex]

    and thus

    [tex]0.99 < \frac{\sin x}{x} < 1.01[/tex]

    The right-hand inequality is true for all [itex]x[/itex], so we only need to consider the left inequality:

    [tex]0.99 < \frac{\sin x}{x}[/tex]

    Now you need to find what range of [itex]x[/itex] satisfies this inequality. I assume you are permitted to do so numerically (using a computer or calculator)?
     
  4. Nov 3, 2009 #3
    Yes, that works. Thank you.
     
  5. Nov 3, 2009 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oops, for the record, I just noticed that I lied when I said that [tex]|\sin(x)/x - 1|[/tex] can equal 0. It can't. But my point that you don't need the [tex]0 < |\sin(x)/x - 1|[/tex] part of the inequality is still correct.
     
  6. Nov 3, 2009 #5
    Ok. :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: More Epsilon Delta Limits
Loading...