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Thread: 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 $\displaystyle \delta$.

    First realize that we have $\displaystyle x_0 - \delta_1 = 8.9401$ and $\displaystyle x_0 + \delta_2 = 9.0601$. From that you can solve for $\displaystyle \delta_1\text{ and } \delta_2$ then take $\displaystyle \delta = \min(\delta_1,\delta_2)$.
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  3. #3
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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 $\displaystyle f(x)-L=3x^2-2x-8$ 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 $\displaystyle f(x)-L=3(x-2)^2+10(x-2)$ and hence you get

    $\displaystyle |f(x)-L|=|3(x-2)^2+10(x-2)|=|x-2|\cdot|3(x-2)+10|<\varepsilon$
    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 $\displaystyle \delta := \min(\varepsilon/M,1)$ 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 $\displaystyle 3x^2- 2x- 4$ really is 4, then $\displaystyle 3x^2- 2x- 4$ must approach 4 as x goes to 2- that means that $\displaystyle 3x^2- 2x- 4- 4= 3x^2- 2x- 8$ must approach 0. And that means that the polynomial $\displaystyle 3x^2- 2x- 8$ must be equal to 0 when x is equal to 2. and that, in turn, means that x- 2 is a factor: $\displaystyle 3x^2- 2x- 8= (x- 2)(3x+ 4)$.
    Now, you have $\displaystyle |x-2||3x+4|< 0.01$ and you want $\displaystyle |x- 2|$ less than something. Of course, as long as $\displaystyle 3x+ 4$ is positive, $\displaystyle |x-2||3x+4|< .001$ is the same as $\displaystyle |x- 2|< \frac{.001}{3x+ 4}$. 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 $\displaystyle |x- 2|< A< \frac{.001}{3x+ 4}$. Now $\displaystyle A< \frac{.001}{3x+ 4}$ is the same (again, assuming 3x+ 4 is positive) as $\displaystyle 3x+ 4< \frac{.001}{A}$ so this reduces to the problem of finding an upper bound on $\displaystyle 3x+ 4$. Obviously we want x close to 0 so it is sufficient to assume -1< x< 1. When x= -1, $\displaystyle 3x+ 4= 3(-1)+ 4= 1$ and when x= 1, 3x+ 4= 3(1)+ 4= 7 so 7 is an upper bound. A= 7, we have $\displaystyle \delta= \frac{.001}{7}$

    (Notice the problem said "find a value of $\displaystyle \delta$". There is no one answer. Taking $\displaystyle \delta= \frac{.001}{7}$ 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 $\displaystyle |x- 9|< \delta$ which is the same as $\displaystyle 9- \delta< x< 9+ \delta$ so it looks like you want $\displaystyle \delta$ to satisfy either $\displaystyle 8.9401< 9- \delta$ or $\displaystyle 9+ \delta< 9.0601$. Find the values of $\displaystyle \delta$ that satisfies those and choose the smaller.
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