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Math Help - Epsilon-Delta Proofs :(

  1. #1
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    Epsilon-Delta Proofs :(

    DELTA-EPSILONS
    1) What is the limit of f(x) = 3x2 - 2x - 4 as x approaches 2?
    Obviously it is 4, now to 1a)
    1a) For epsilon = 0.01, find a value of delta so that f(x) is closer to the limit than epsilon. Apply the same for when epsilon = 0.001.

    2) What is the limit of f(x) = sqrt(x) as x approaches 9?
    Obviously it is 3, now to 2a)
    2a) For epsilon = 0.01, find a value of delta so that f(x) is closer to the limit than epsilon. Apply the same for when epsilon = 0.001.

    Now, I actually started both #1 and #2. My work, respectively:

    1a) Let epsilon = 0.01
    • |f(x) - L| < epsilon
    • |3x^2 - 2x - 4 - 4| < 0.01
    • -0.01 < 3x^2 - 2x - 8 < 0.01

    Now, where do I go from here?
    2a) Let epsilon = 0.01
    • |f(x) - L| < epsilon
    • |sqrt(x) - 3| < 0.01
    • -0.01 < sqrt(x) - 3 < 0.01
    • 2.99 < sqrt(x) < 3.01
    • 2.992 < sqrt(x) < 3.012
    • 8.9401 < x < 9.0601

    Now, where do I go from here?
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  2. #2
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    From here you want to determine \delta.

    First realize that we have x_0 - \delta_1 = 8.9401 and x_0 + \delta_2 = 9.0601. From that you can solve for \delta_1\text{ and } \delta_2 then take \delta = \min(\delta_1,\delta_2).
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  3. #3
    Super Member Failure's Avatar
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    1. a)

    Quote Originally Posted by Shaidester View Post
    DELTA-EPSILONS
    1) What is the limit of f(x) = 3x2 - 2x - 4 as x approaches 2?
    Obviously it is 4, now to 1a)
    1a) For epsilon = 0.01, find a value of delta so that f(x) is closer to the limit than epsilon. Apply the same for when epsilon = 0.001.





    1a) Let epsilon = 0.01
    • |f(x) - L| < epsilon
    • |3x^2 - 2x - 4 - 4| < 0.01
    • -0.01 < 3x^2 - 2x - 8 < 0.01

    Now, where do I go from here?
    You can develop the polynomial f(x)-L=3x^2-2x-8 in terms of x-2, by substituting (x-2)+2 for x in f(x)-L and then expanding but keeping the term (x-2) together.
    This gives f(x)-L=3(x-2)^2+10(x-2) and hence you get

    |f(x)-L|=|3(x-2)^2+10(x-2)|=|x-2|\cdot|3(x-2)+10|<\varepsilon
    Next you need to find an upper bound for the second factor on the left of the above inequality. To this end you must first limit the range for x (otherwise that second factor would not be bound). For example, you can limit x to the intervall [2-1,2+1]=[1,2]. Now figure out that upper bound M for the second factor on the left for x from this intervall and then divide the entire inequality by that upper bound. You get that \delta := \min(\varepsilon/M,1) does the job.
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  4. #4
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    Quote Originally Posted by Shaidester View Post
    DELTA-EPSILONS
    1) What is the limit of f(x) = 3x2 - 2x - 4 as x approaches 2?
    Obviously it is 4, now to 1a)
    1a) For epsilon = 0.01, find a value of delta so that f(x) is closer to the limit than epsilon. Apply the same for when epsilon = 0.001.

    2) What is the limit of f(x) = sqrt(x) as x approaches 9?
    Obviously it is 3, now to 2a)
    2a) For epsilon = 0.01, find a value of delta so that f(x) is closer to the limit than epsilon. Apply the same for when epsilon = 0.001.

    Now, I actually started both #1 and #2. My work, respectively:

    1a) Let epsilon = 0.01
    • |f(x) - L| < epsilon
    • |3x^2 - 2x - 4 - 4| < 0.01
    • -0.01 < 3x^2 - 2x - 8 < 0.01

    Now, where do I go from here?
    Note that if the limit of 3x^2- 2x- 4 really is 4, then 3x^2- 2x- 4 must approach 4 as x goes to 2- that means that 3x^2- 2x- 4- 4= 3x^2- 2x- 8 must approach 0. And that means that the polynomial 3x^2- 2x- 8 must be equal to 0 when x is equal to 2. and that, in turn, means that x- 2 is a factor: 3x^2- 2x- 8= (x- 2)(3x+ 4).
    Now, you have |x-2||3x+4|< 0.01 and you want |x- 2| less than something. Of course, as long as 3x+ 4 is positive, |x-2||3x+4|< .001 is the same as |x- 2|< \frac{.001}{3x+ 4}. But that won't do because you need a NUMBER on the right, not a function of x.

    It would be sufficient to find a number "A" such that |x- 2|< A< \frac{.001}{3x+ 4}. Now A< \frac{.001}{3x+ 4} is the same (again, assuming 3x+ 4 is positive) as 3x+ 4< \frac{.001}{A} so this reduces to the problem of finding an upper bound on 3x+ 4. Obviously we want x close to 0 so it is sufficient to assume -1< x< 1. When x= -1, 3x+ 4= 3(-1)+ 4= 1 and when x= 1, 3x+ 4= 3(1)+ 4= 7 so 7 is an upper bound. A= 7, we have \delta= \frac{.001}{7}

    (Notice the problem said "find a value of \delta". There is no one answer. Taking \delta= \frac{.001}{7} works but so would any smaller number and possibly some larger numbers.)

    2a) Let epsilon = 0.01
    • |f(x) - L| < epsilon
    • |sqrt(x) - 3| < 0.01
    • -0.01 < sqrt(x) - 3 < 0.01
    • 2.99 < sqrt(x) < 3.01
    • 2.992 < sqrt(x) < 3.012
    • 8.9401 < x < 9.0601

    Now, where do I go from here?
    Well, you want |x- 9|< \delta which is the same as 9- \delta< x< 9+ \delta so it looks like you want \delta to satisfy either 8.9401< 9- \delta or 9+ \delta< 9.0601. Find the values of \delta that satisfies those and choose the smaller.
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