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Delta-Epsilon Arguments

  1. Dec 15, 2013 #1
    1. The problem statement, all variables and given/known data

    For the given limit and the given ε, find the largest value of δ that will guarantee the conclusion of the statement.

    f(x)=8x+15, ε=1, lim f(x) = 95
    x→10

    2. Relevant equations

    So the statement that is on the worksheet says:

    Let L be a real number and let f be a function defined on an open interval containing c, but not necessarily defined at c.

    the statement lim f(x) = L means that for all ε > 0, there exists δ > 0 such that if 0 < |x-c| < δ then |f(x) - L| < ε
    ....................x→c



    3. The attempt at a solution

    This is what I got and I not even sure it's correct.


    |(8x+15) - 95| < 1
    |8x - 80| < 1

    now what?
     
  2. jcsd
  3. Dec 15, 2013 #2

    Dick

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    Well, that's an ok start. What's supposed to be less than δ for that to be true?
     
  4. Dec 15, 2013 #3
    Do I solve the equation?

    So 8x < 81

    x < 81/8

    81/8 is the answer?
     
  5. Dec 15, 2013 #4

    Dick

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    No, that's not right. You want |8x - 80| < 1 whenever |x-10|<δ. What's the largest value of δ you can choose that will make that work?
     
  6. Dec 15, 2013 #5
    I can not wrap my mind around this. Would it be 10?
     
  7. Dec 15, 2013 #6

    Dick

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    No, you may be a little shakey on how to work with absolute value. If |x-10|<10 then the solution to that is all numbers 0<x<20. Do you see why? All of those values of x don't work with |8x-80|<1, do they? x=10 works. x=11 doesn't work. Which values of x do work? Think about this a little more. Maybe something will click.
     
  8. Dec 15, 2013 #7
    If |x-10|< .125 then the solutions would be from 0<x<10.125 and that works?
     
  9. Dec 15, 2013 #8

    Dick

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    That's real progress! The solutions are actually 10-0.125<x<10+0.125. But you've got it. How did you get 0.125?
     
  10. Dec 15, 2013 #9
    Solved |8x - 80| < 1. and that is 10.125. Put 10.125 into |x-10| < δ.

    So the largest value of δ that would guarantee the problem would be 10.125?
     
  11. Dec 15, 2013 #10

    Dick

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    You said |x-10|<0.125. Doesn't that make δ=0.125? Look at your limit definition. And you are still being a little sloppy on the absolute value solutions. You solved 8x-80=1, that's really only a part of solving |8x-80|<1. Review absolute value inequalities, ok? |x-a|<b means a-b<x<a+b. It means the distance between x and a is less than b. Here's another way to go once you've got the absolute value stuff. |8x-80|=8|x-10|. Agree? So if |8x-80|<1 and |x-10|<δ what's the largest value of δ that will work?
     
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