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Math Help - Calculus, limits

  1. #1
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    Calculus, limits

    For the limit below, find values of δ that correspond to the ε values.




    ε = .01 what is the δ rounded to five decimal places. Can anyone please explain to me how to do this problem? I know that the easiest way is to plug it into your graphing calculator and trace the line, but my TI-83 Plus is not accurate enough to give me any answer other than 1. Thank you!
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  2. #2
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    Quote Originally Posted by fruitkate21 View Post
    For the limit below, find values of δ that correspond to the ε values.




    ε = .01 what is the δ rounded to five decimal places. Can anyone please explain to me how to do this problem? I know that the easiest way is to plug it into your graphing calculator and trace the line, but my TI-83 Plus is not accurate enough to give me any answer other than 1. Thank you!
    I'm guessing that what you mean is to find the largest positive delta such that 3.99 < 6 + x - 3x^3 < 4.01 is true for all values of x in the interval 1 - \delta < x < 1 + \delta.
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    yes, that is what i mean, do you know how to do the math that goes along with it?
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  4. #4
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    What I would do is this: First determine that your function will be decreasing on the interval (3.99, 4.01) by taking the derivative and noting that it is negative. Then solve the equations 6 + x - 3x^3 = 4.01 and 6 + x - 3x^3 = 3.99. The first will give you a value of x less than 1 and the second will give you a value of x greater than one. Since the function is decreasing on the interval (3.99, 4.01), the Extreme Value Theorem applies to say that these values of x yield the minimum and maximum values of f(x) on the interval. Now, you choose which of those values of x results in the smaller delta from 1.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fruitkate21 View Post
    yes, that is what i mean, do you know how to do the math that goes along with it?
    i assume you know the definition you should be working with. i will do 95% of the problem for you and leave you to finish of

    we want |f(x) - 4| < \epsilon

    \Rightarrow |6 + x - 3x^3 - 4| < \epsilon

    \Rightarrow |2 + x - 3x^3| < \epsilon

    \Rightarrow |3x^3 - x - 2| < \epsilon

    \Rightarrow |3x^2 + 3x + 2||x - 1| < \epsilon

    Now, when x is "near" 1, we have:

    8|x - 1| < \epsilon

    \Rightarrow |x - 1| < \frac {\epsilon}8

    now finish up
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  6. #6
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    I am having problems solving the equation 6 + x - 3x^3 = 4.01, would you mind helping me figure out how to do that?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fruitkate21 View Post
    I am having problems solving the equation 6 + x - 3x^3 = 4.01, would you mind helping me figure out how to do that?
    yes, that would be a pain. you would probably need Newton's method.

    it would be easier to solve it were it 4 instead of 4.01 or 3.99, which is what the method i used requires. did you get it?
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  8. #8
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    Jhevon's method is probably easier than mine, since solving cubic equations is difficult. And yes, you would use Newton's method to approximate the answer to five decimal places of accuracy. Either that or plug it into your calculator's equation solver.
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    i assume you know the definition you should be working with. i will do 95% of the problem for you and leave you to finish of

    we want |f(x) - 4| < \epsilon

    \Rightarrow |6 + x - 3x^3 - 4| < \epsilon

    \Rightarrow |2 + x - 3x^3| < \epsilon

    \Rightarrow |3x^3 - x - 2| < \epsilon

    \Rightarrow |3x^2 + 3x + 2||x - 1| < \epsilon

    Now, when x is "near" 1, we have:

    8|x - 1| < \epsilon

    \Rightarrow |x - 1| < \frac {\epsilon}8

    now finish up
    I know this is a old post but I need help on this problem. Can explain to me how you got
    \Rightarrow |2 + x - 3x^3| < \epsilon
    to this step:
    \Rightarrow |3x^3 - x - 2| < \epsilon
    \Rightarrow |3x^2 + 3x + 2||x - 1| < \epsilon
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by yangx View Post
    I know this is a old post but I need help on this problem. Can explain to me how you got
    \Rightarrow |2 + x - 3x^3| < \epsilon
    to this step:
    \Rightarrow |3x^3 - x - 2| < \epsilon
    \Rightarrow |3x^2 + 3x + 2||x - 1| < \epsilon
    |2+x-3x^3|=|-(3x^3-x-2)|=|-1||3x^3-x-2|=|3x^3-x-2|
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yangx View Post
    I know this is a old post but I need help on this problem. Can explain to me how you got
    \Rightarrow |2 + x - 3x^3| < \epsilon
    to this step:
    \Rightarrow |3x^3 - x - 2| < \epsilon
    Here, I simply changed all the signs. (|a| = |-a|, it makes no difference). this was more for aesthetic reasons and ease of factoring, which I did in the next step

    \Rightarrow |3x^2 + 3x + 2||x - 1| < \epsilon
    here is where i factor the left hand side of the previous step. i divided 3x^3 - x - 2 by x - 1 and i got 3x^2 + 3x + 2, and so i could factor it as you see.

    how did i know to divide by x - 1? it is what we needed to factor out. it is our |x - x_0| < \delta term. here, x_0 = 1



    ...in retrospect, i would have actually done this a bit differently..., first letting |x - 1| < 1 or something as an initial delta bound.
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