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Homework Help: 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.


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


    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


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    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]


    [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


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    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. :)
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