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Epsilon Delta Limits

  1. Jan 22, 2008 #1
    Could someone please give me a walkthrough of the following question(and answer)??
    I really can't understand it....

    lim x^2 = 9
    x->3

    if 0<|x-c|<delta then |f(x) - L|< epsilon
    so... x^2 - 9 = (x+3)(x-3)
    |x^2 - 9| = |x+3||x-3|


    Here's the problem.The book states:

    An estimate of what??And how do I know when I must estimate??I'm guessing that it doesnt matter whether (x+3) or (x-3) is chosen due to the absolute value brackets..?
    anyway:
    if |x-3|<1 then 2<x<4 (Choosing a delta....now x lies between 2 and 4.)
    |x+3|<=|x|+|3|= x+3<7 (this would be: c-delta<x<c+delta,correct?) **How do I know when to implement that ??**Why are they now using |x+3| instead of (x-3)?!?

    if |x-3|<1 then |x^2 -9|< 7|x-3|
    I understand that this solution basically states delta = 1/7 epsilon??And I know that the 7 is acquired through the c-delta<x<c+delta formula/method but I'm still too lost!!

    please clarify guys!
     
    Last edited: Jan 22, 2008
  2. jcsd
  3. Jan 22, 2008 #2

    HallsofIvy

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    Good so far. You want [itex]|x+3||x-3|< \epsilon[/itex] and you have to make sure that is true when [itex]|x-3|< \delta[/itex]. Notice that one has |x+3| and the other doesn't!


    An estimate of how large |x+3| is close to 3. And it certainly does matter "whether (x+3) or (x-3) is chosen". You are taking the limit as x-> 3, not -3! You want get [itex]|x-3|< \delta[/itex], not [itex]|x+3|< \delta[/itex].


    You want [itex]|x-3||x+3|< \epsilon[/itex] whenever [itex]|x-3|< \delta[/itex] and the question is, how should you choose [itex]\delta[/itex].
    You could try writing [itex]|x- 3|< \epsilon/|x+3|[/itex] but that depends on x, it is not a constant. That's where the "estimate for |x+3|" comes in. You need to replace |x+3| by a constant, K, and be sure that K> |x+3| so that [itex]\epsilon/K< \epsilon/|x+3|[/itex].

    Since we are going to be "close to 3" anyway, suppose |x-3|< 1- that is, -1< x-3< 1. Adding 3 to both sides, 2< x< 4 and, adding another 3, 5< x+3< 7 so if |x-1|< 1, |x+3|< 7. Then 1/7< 1/|x+3| (notice how the inequality swaps direction when we put the numbers in the denominator: 2< 3 so 1/3< 1/2). Since 1/7< 1/|x+3|, [itex]\epsilon/7< \epsilon/|x+3|[/itex]. We can choose [itex]\delta[/itex] equal to the smaller of 1 and [itex]\epsilon/7[/itex] so they are both true.

    If [itex]|x-3|< \delta< \epsilon/7< \epsilon/|x+3|[/itex], then, multiplying both sides by the positive number |x+3|, [itex]|x-3||x+ 3|= |x^2- 9|< \epsilon[/itex].

    (Of course, the "|x-3|< 1" is just convenient. |x-3|< any fixed positive number would work as well.)
     
  4. Jan 22, 2008 #3
    OK I'm going to print this out in the morning and get back to you on this one.(in the South African morning that is)Thanks...
     
  5. Jan 23, 2008 #4
    OK,I had a look last night.I had problems with the following:

    I didn't realise that the value x was approaching(in this case 3),had any other function than to be plugged in |x-c|<delta.So I learned something there.But :
    I don't know where |x-1| came from....

    I can see that Epsilon is 7 times bigger than delta(in Salas) but I don't quite understand this method of making the inequality into a fraction...

    And just to check:
    0<|x-c|<delta is equivalent to c-delta<x<c+delta and can be interchanged at any time??
     
    Last edited: Jan 23, 2008
  6. Jan 23, 2008 #5

    HallsofIvy

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    That was a typo. I meant, |x-3|, of course.

    If 0< a< b, (and neither is 0) then "taking the reciprocal" reverses the inequality: 1/a> 1/b. You can do that step by step: first divide both sides by the positive number a: 1< b/a, and then divide both sides by the positive number b: 1/b< 1/a which is the same as 1/a> 1/b.

    Yes, that follows from the definition of absolute value. If x>= c, then x-c is non-negative so |x-c|= x- c: x- c< delta => x< c+ delta. If x< c, then x- c is negative so |x- c|= -(x-c): -(x-c)< delta=> x-c> -delta=> x> x- delta.
     
  7. Jan 23, 2008 #6
    Ok I attempted a very similar problem to try and get a feel for this,although I have gotten a bit stuck.This is what I've done:

    question: lim x^2 =4
    x->2
    my answer:

    |x-2|<delta
    |x^2 - 4|<epsilon
    |x-2||x+2|<epsilon
    Let |x+2|<1
    -1<x+2<1
    -3<x<-1 It's gone negative here...Doesn't seem right.
    -5<x-2<-3
    |x-2|<-3

    couldn't i assume that if |x-2|<-3 then |x+2|<1???


    I would have assumed that if |x-1|<1 then |x+3|<4...(by adding 4 to |x-1| to equate it with |x+3| ?)
     
    Last edited: Jan 23, 2008
  8. Jan 23, 2008 #7

    HallsofIvy

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    You've got this backwards. Since the limit is take at x= 2, you want x close to 2, so x- 2 (not x+2) close to 0. For example, |x-2|< 1.

    |x-2| is NOT "< -3", absolute value is never negative. Saying that -5< x-2< -3 tells you, first, that x-2< 0! In that case, |x-2|= -(x-2). Multiplying each part of the inequality by the negative number -1, 5> -(x-2)> 3 or 3< |x-2|< 5. |x-2|< 5.
    Actually, you can get that directly from -5< x-2< -3. The larger of the two absolute values, 5 and 3, is 5: |x-2|< 5.

    However, you should be thinking: Take x close to 2: say |x- 2|< 1. Then -1< x- 2< 1 so, adding 4 to each part, 3< x+ 2< 5. |x+2|< 5. Then |x-2||x+2|< 5|x-2|< [itex]\epsilon[/itex] so we need |x-2|< [itex]\epsilon/5[/itex]. Take [itex]\delta[/itex] to be the smaller of [itex]\epsilon/5[/itex] and 1.

    Is this a different example? If |x-1|< 1, then -1< x- 1< 1 so 3< x+ 3< 5. |x+3|< 5.
     
  9. Jan 24, 2008 #8

    HallsofIvy

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    The function does not have to "pass through both corners". But you want to make the estimate as accurate as possible so you construct the rectangle that way. Where ever your "given" horizontal lines cross the graph, those are the corners of your rectangle.
     
    Last edited: Jan 24, 2008
  10. Jan 24, 2008 #9
    OK ok...I think for now this is as good as I'm going to get it!

    **But be warned!!I will be spamming the forums with difference quotient problems on Monday.Let some of the other mods have a turn(-8**

    Thanks Halls
     
    Last edited: Jan 24, 2008
  11. Feb 7, 2008 #10
    additional question

    Quote: Multiplying each part of the inequality by the negative number -1, 5> -(x-2)> 3 or 3< |x-2|< 5. |x-2|< 5.

    Question: I am having trouble understanding the concept behind a variation of this... for example, I do understand getting into the format of -1 < (x-2) <1 then you add 2 to both sides and then you add another two to get into the format of (x+2). However what happens if you are trying to get into another format because the original equation is different and end up, for example, with
    -3<(x+2)<5.... meaning what happens when one side is negative and the other isn't.. is there ever a case where you end up taking the smaller of the 2 absolute values or is it always the case that it is the larger absolute value that you end up being "less than." If there is a negative value on either side, do you always multiply by -1 which could then lead to the other side being negative?

    I appreciate any help with this concept.

    Thank you,

    jason
     
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