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Homework Help: Why use stipulations in Epsilon Delta Proofs?

  1. Dec 28, 2012 #1
    1. The problem statement, all variables and given/known data
    When constructing an Epsilon Delta proof, why do we need to make a stipulation? For example, in most proofs for limits of quadratic functions, it is stipulated, for example, that δ≤1. Why is this needed anyway?

    This is my thought process for a quadratic:

    Prove that lim(x -->3) x^2 = 9

    2. Relevant equations

    3. The attempt at a solution
    For every ε > 0 there is δ > 0 such that |x^2 - 9| < ε whenever 0 < |x-3| < δ

    |x+3||x-3| < ε

    We will try to find a number K such that
    |x+3||x-3| < K|x-3| < ε

    Which will give us |x+3| < K and |x-3| < ε/K

    This is the point at which it is generally stipulated that δ ≤ 1.

    Instead, I used the original statement from above: |x-3| < δ

    -δ + 3 < x < δ + 3
    -δ + 6 < x + 3 < δ + 6

    Now, if x+3 = |x+3|, then x+3 must be ≥0.
    Therefore, -δ + 6 ≥ 0 --> δ≤6

    -δ + 6 < |x + 3| < δ + 6 (with the condition that δ≤6)

    Since |x+3| < K and |x + 3| < δ + 6 --> K ≥ δ + 6

    We have found K for any δ > 0 we choose, so we have shown that |x+3| < K
    for some value K, and so,

    |x-3| < ε/K

    Thus, we can say that δ ≤ ε/K to guarantee that this is true


    if 0 < |x-3| < ε/K then |x^2 - 9| < ε
    We assumed that |x-3| < ε/K and that |x+3| < K

    |x^2 - 9| = |x+3||x-3|
    < K|x-3|
    < K(ε/K)
    < ε

    My question is, did I do something wrong by not stipulating that δ is less than or equal to some value? Is my method logically correct?
    Last edited: Dec 28, 2012
  2. jcsd
  3. Dec 28, 2012 #2


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    There was a much easier way to do your proof :

    [itex]|x^2-9| = |x-3||x+3| < \delta |x+3|[/itex]

    Now using the triangle inequality and a little trick for a small calculation, observe that :

    [itex]|x+3| = |x-3+6| ≤ |x-3| + 6 < \delta + 6[/itex]

    So continuing you could say :

    [itex]\delta |x+3| < \delta(\delta + 6)[/itex]

    Okay, now for your 'stipulation'. We now have [itex]|x^2-9|[/itex] completely in terms of delta which is what we set out to do for these kinds of proofs.

    So we want to find δ in terms of ε now by satisfying :

    [itex]\delta(\delta + 6) ≤ ε[/itex]

    Notice that this is not so easy to manipulate, which is why we sometimes allow delta to be relatively small and bounded as to make the calculation easier. If we don't then this is much more difficult than it looks.

    Hence we can choose delta as small as we like, even as small as 0.000000001, but 1 is used for convenience since it is very easy to work with. So suppose we have δ ≤ 1.

    What happens to [itex]\delta(\delta + 6)[/itex]? Could you tell me why ε/K does not always work? ( Hint : Think about how big your neighborhood is ).
  4. Dec 28, 2012 #3


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    Our goal is to find a δ such that every number close to 3 (i.e. 0 < |x-3| < δ) causes x^2 to be sufficiently close to 9 (i.e. within ε).

    Our approach to the problem is to start by studying the end result: we want
    [tex]|x^2 - 9| < \epsilon[/tex]

    which we simplify to
    [tex]|x-3||x+3| < \epsilon[/tex]

    How can we ensure [itex]|x-3||x+3|[/itex] is small for x close to 3? Well, we can look at each factor individually.

    The closer x is to 3, the closer |x-3| is. Our simplest tool to make this small is to pick δ very small.

    If x is very close to 3, however, |x+3| isn't very small: it's around 6. |x+3| would be small if x were near -3, but we can't pick δ in a way to force x to be near -3.

    However, there is a solution: because we have a way to make |x-3| small, all we have to do is prevent |x+3| from becoming big.

    Fortunately, this is pretty easy: as I mentioned above, if x were near 3, then |x+3| would be near 6. To make a precise and simple statement we can work with, we can simply insist that [itex]|x+3| < 7[/itex] by requiring [itex]2 < x < 4[/itex], which we can do by requiring [itex]\delta \leq 1[/itex].

    From the problem we are trying to solve, [itex]x[/itex] is forced to be a number such that |x-3| is smaller than δ. By our additional choice to consider only [itex]\delta \leq 1[/itex], we know |x+3| is smaller than 7. Thus

    If [itex]\delta \leq 1[/itex] then [itex]|x+3||x-3| < 7 \delta[/itex]​

    If we also require [itex]\delta \leq \epsilon/7[/itex] , then we can ensure [itex]|x^2 - 9| < \epsilon[/itex]

    This analysis has put two conditions on δ such that, if satisfied, the thing we are trying to prove is true. So the problem now boils down to finding a solution to the system of inequalities

    [tex]0 < \delta \qquad \qquad \delta \leq 1 \qquad \qquad \delta \leq \epsilon / 7[/tex]

    If this system didn't have a solution, we'd have to try harder (note the analysis we made is not sufficient to prove the limit doesn't exist if this system has no solutions). Fortunately, it does have solutions, such as

    [tex]\delta = \min(1, \epsilon / 7)[/tex]
  5. Dec 28, 2012 #4


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    I wanted to add this sort of thing is typical in analysis, and to some extent calculus. When you want to understand how things behave as x approaches 3, a standard technique is to try and separate things into two types of things:
    • Things that have a lot of effect
    • Things that don't have much effect

    The trick for the first type is to try and express these things as simply as possible, even at the cost of making the rest of the expression more complex. In this case, the important part was |x-3|.

    For the parts that don't have much effect, you can usually make very crude simplifications that greatly reduce their complexity: in this case, [itex]x+3[/itex] is uninteresting as x approaches 3, so we want to simply replace it with 6 to simplify the problem.

    So we want to understand [itex]|x^2 - 9|[/itex] as behaving the same as the simpler expression [itex]|6(x-3)|[/itex].

    And then, you convert the idea into proof. We already understand how to find the limit of [itex]|6(x-3)|[/itex]. There is the added complication of ensuring that [itex]|6(x-3)|[/itex] is a sufficiently good approximation for [itex]|x^2 - 9|[/itex], which is where the stipulation comes from.

    It might be interesting to note that
    [tex]\left(6(x-3) \right) - \left( x^2 - 9 \right) = -(x-3)^2[/tex]
  6. Dec 30, 2012 #5


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    If we just consider your proof below ...

    You haven't actually defined how you find K for any particular ε.
    If you want to find δ as a function of ε, which is valid for any value of ε, where ε>0, that can be done. One such function is :
    [itex]\displaystyle \delta=\sqrt{9+\varepsilon}-3\ .[/itex]​
    (Notice that if ε=7, then δ=1 ... i.e. δ = ε/7 .)

    Showing that this δ fulfills the necessary condition, may be quite a bit more difficult than it is to show that δ = min(1, ε/7) fulfills the necessary condition.
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