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Trying to find delta in a limit involving an inequality

  1. Sep 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Find δ (which is the input tolerance):

    lim (9 - 4x2)/(3 + 2x2) = 6
    x→-1.5

    2. Relevant equations

    |f(x) - L| < [itex]\epsilon[/itex]
    |x - a| < [itex]\delta[/itex]

    3. The attempt at a solution

    lim (9 - 4x2)/(3 + 2x2) = 6
    x→-1.5

    I need to get to:
    |x-(-1.5)| < [itex]\delta[/itex]
    =
    |x + 1.5)| < [itex]\delta[/itex]


    So:

    |[(9 - 4x2)/(3 + 2x2)] - 6| < ε

    |[(3 - 2x)(3 + 2x)/(3 + 2x)] - 6| < ε

    |-2x - 3| < ε

    |-2(x + 1.5)| < ε

    |x + 1.5| < -ε/2

    Therefore, δ = -ε/2

    But when you divide an inequality equation by -2, aren't you supposed to switch the operator?

    But the answer |x + 1.5| < -ε/2 = [itex]\delta[/itex] is what I wanted.

    Thanks!
     
    Last edited: Sep 2, 2013
  2. jcsd
  3. Sep 2, 2013 #2

    pasmith

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    Here is your error:
    [tex]|{-2}(x + 1.5)| = 2|x + 1.5|[/tex]
     
  4. Sep 2, 2013 #3
    Thank you pasmith for your answer!

    But if |-2(x + 1.5)| = 2|x + 1.5|

    then does it matter? The answers above will both be positive, right? I realize that I probably am not seeing something.
     
  5. Sep 2, 2013 #4

    Zondrina

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    How about you try applying the triangle inequality to |x + 1.5| .
     
  6. Sep 2, 2013 #5
    Hi again Zondrina!

    By triangle inequality, do you mean:

    0 < |x + 1.5| < δ ??

    I am not sure what to do.
     
  7. Sep 2, 2013 #6
    Ok, I am going to leave the restaurant where I am posting this at and getting wifi to another place. I will be there in about 45 minutes. So I am not being rude if I don't reply to any responses in 45 minutes or so!
     
  8. Sep 2, 2013 #7

    Zondrina

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    Hint : ##|x + 1.5| = |x - 1.5 + 3| ≤ \space ?##
     
  9. Sep 2, 2013 #8
    Hi again,

    I am not trying to be dense, but I don't even know how to proceed with the clue.

    We only did one limit proof with delta and epsilon in class. We also had two other homework problems involving these, but neither one involved dividing or multiplying by a negative with the inequality, and they worked out as intended.

    I know that if you solve an inequality by dividing or multiplying with a negative number, you have to reverse the operator, either from < to > or from > to <. But in the original problem, this would give me just the opposite answer I was looking for.
     
  10. Sep 2, 2013 #9

    Zondrina

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    The triangle inequality : ##|x + y| ≤ |x| + |y|, \space \forall x \in ℝ##

    So if we applied it to ##|x - 2 + 4|## for example, you would get : ##|x - 2 + 4| ≤ |x - 2| + 4##.
     
  11. Sep 2, 2013 #10
    Thanks!

    But isn't |x + y| = |x| + |y|? Not [itex]\leq[/itex] ?
     
  12. Sep 2, 2013 #11

    Zondrina

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    As a quick counter-example to your claim, take ##x=2## and ##y=-1##.
     
  13. Sep 2, 2013 #12
    The δ I am getting is

    |-2(x + 1.5)| < ε

    |x + 1.5| > -ε/2

    but I need to get |x + 1.5| < something
     
  14. Sep 2, 2013 #13
    And of course, you are right. I was thinking of a triangle, since it is called the "triangle inequality". So I was just thinking of lengths, which have to be positive.

    Also, I admit that I am very slow to catch on, but I am still not sure what the triangle inequality has to do with my original problem.

    I am trying to convert the operator from ">" to "<" so that | x + 1.5| < δ
    All I have now is |x + 1.5| > -ε/2
    I am so close, but I don't know how to get it in the format I want.
     
  15. Sep 2, 2013 #14

    Zondrina

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    Apply the triangle inequality to my prior hint : ##|x + 1.5| = |x - 1.5 + 3| ≤ \space ?##

    Hint : ##|x + 1.5| = |(x - 1.5) + 3| ≤ \space ?##
     
  16. Sep 2, 2013 #15
    So:

    |x - 1.5 + 3| [itex]\leq[/itex] |x - 1.5| + |3|

    Applied to my original problem:

    |-2x - 3| < ε

    |-2(x + 1.5)| < ε

    |x + 1.5| > -ε/2

    |x - 1.5 + 3| [itex]\leq[/itex] |x - 1.5| + |3| > -ε/2

    |x - 1.5| > -ε/2 - 3

    But now I both need to change the operator from ">" to "<" and change |x - 1.5| to |x + 1.5|.
     
  17. Sep 2, 2013 #16

    Zondrina

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    I just noticed something, which I would like to apologize for. I didn't see that tiny negative sign next to the 1.5, so it changes what I said earlier a bit. I also noticed it was linear so what I said earlier is completely irrelevant.

    So here's what you reduced it to earlier which is correct :

    ##2|x + 1.5|##

    Now you know ##|x + 1.5| < δ## so that ##2|x + 1.5| < 2δ##.

    You can now find ##δ## in terms of ##ε## like so :

    ##2δ ≤ ε \Rightarrow δ ≤ \frac{ε}{2}##.

    Once again, sorry about earlier.
     
  18. Sep 2, 2013 #17
    First of all, there is no need to apologize! You are helping people out on this board, which is the main thing; and that is very laudable! I am thrilled to even get any help. So no worries.

    Second, the equation I calculated was:

    |-2(x + 1.5)| < ε

    but you wrote above:

    2|x + 1.5|

    I guess both expressions are equal, after I tried some numbers. But I don't know the rule that allows you to pull a number out of an absolute value equation and change the sign. Is there a rule about this that I should learn, or maybe it's just obvious (to everyone except me!! :tongue: )

    The above is what confused me in my original question I guess.
     
  19. Sep 2, 2013 #18

    Zondrina

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    There is indeed a rule. ##|ab| = |a||b|, \forall a, b \in ℝ##

    So in the case of ##|-2(x+1.5)|##, you have ##a = -2## and ##b = x + 1.5##.
     
  20. Sep 2, 2013 #19
    Thank you! Now it looks obvious to me, but it wasn't until you showed me the rule.
     
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