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Critique My proof:

  1. Jan 4, 2012 #1
    I am going to make this my official thread directly for my proof questions so I dont have to keep making new threads. And sorry if my questions all seem redundant because I realize that they do but I really want to better my proof writing and understand how to write proofs so thank you for everyone who has tryed to help.

    Anyways first proof:
    Im starting off with something easy because right now I am more focused on concentraiting on the correct syntax and verbage and what not that I should use.

    I want to show that f(x) → a^2 near a.

    Proof:
    For every ε>0 there is a δ>0 such that for all x, excluding x = a, if |x-a|< δ then
    |f(x) - a^2|< ε
    I am starting my proof off with a definition which states important facts namely ε and δ must be greater than zero for ALL x and at the same time explaining what must be true if we expect the limit to exist. In general is this how proofs are normally started by briefly explaining what must be true?

    |f(x) - a^2| < ε = |x^2 - a^2| < ε = |x+a||x-a| < ε.
    For something as easy as this, is it necessary to go step by step from simplifying |f(x) - a^2| to |x+a||x-a|?

    |x| - |a| ≤|x-a|< δ1, |x|<δ1 + |a|, |x+a|≤|x| + |a|< δ1 + 2|a| and consequently |x+a||x-a|<|x+a|(2|a|+δ1).
    Do I have to add an explanation of why I started the new line with |x| - |a| ≤|x-a|< δ1?

    Thus for |x^2 - a^2|<ε we have |x-a| < ε/(2|a| + δ1) for |x-a| = min(δ1,ε/(2|a| + δ1).


    Anyways some suggestion on how to better improve writing proofs would be appreciated
     
  2. jcsd
  3. Jan 4, 2012 #2
    In these ε-δ-proofs you often find the suitable δ after going through the proof without knowing it, i.e. you find an inequality binding ε and δ, solve for δ and assume this holds from the beginning of your proof. It's hard to explain, but you'll see it in the end.

    At this part you should list your assumptions etc. for instance: for any ε>0 choose δ=... (you'll find it in the proof). Now suppose |x-a|<δ ...
    I don't think it is necessary to go step by step, but it "looks" like your starting from what you are about to prove.
    I prefer to just go on without ε for a while:

    |f(x) - a^2| = |x^2 - a^2| = |x+a||x-a| < |x+a|δ.
    Just mention that "to estimate |x|, we notice that..."
    For this part, I would like to estimate δ:

    |f(x) - a^2| = |x^2 - a^2| = |x+a||x-a| < |x+a|δ < δ(δ+2|a|).

    Now make an extra estimate: δ <= 1; if it is not, then we reduce it to such. Notice we have to remember this in the final choice of δ.

    δ(δ+2|a|) <= δ(1+2|a|)
    Now we just solve the inequality δ(1+2|a|) <= ε <=> δ <= ε/(1+2|a|) (There are strict inequalities in the proof, so equality is allowed.)
    So we conclude: For any ε > 0 we choose δ = min{ ε/(1+2|a|) , 1}. You can make the guess in the beginning that this choice suffices and then show that it actually does.
     
  4. Jan 4, 2012 #3

    SammyS

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    This theorem requires a somewhat different sort of proof than the theorem in your thread: "Not understanding simple proof". Also, in my reply to that thread, all of the text displayed in red is to help you understand some of the elements of that proof. They would not usually be included in a formal proof.

    This may be an oversight on your part, but you didn't state anywhere in this thread what the definition of f(x) is, although a person reading the original post can figure what you probably intended it to be.

    So, I take it that you want to prove that f(x) → a2 for values of x near a, for the function, f(x) = x2 .

    This can be stated in briefer form as:
    Prove: [itex]\displaystyle \lim_{x\,\to\,a}\,x^2=a^2\,.[/itex]​

    What is it that you need to prove?
    You need to prove the following.

    For an arbitrary ε > 0, there exists δ > 0, such that for any x that satisfies 0 < |x-a| < δ, it's true that |x2 - a2| < ε .​

    It may be helpful for you to see how this works for two or three different values of a: 0, 3, -2. For each of these, try a few values for ε, perhaps, ε = 1, ε = 1/16 , ε = 0.01. For each of these combinations, find a suitable δ .
    This exercise is not necessarily needed for this proof in specific. But, hopefully it will help you understand some of the interplay between δ and ε in the definition of the limit of a function.​

    The way these proofs are usually started is:

    Let ε > 0.

    Then you define a δ. ( δ usually some expression which depends upon ε. The way you find what will work for δ is not usually shown in the proof. However, this is where the lion's share of the work is when you do a proof. )

    After you state what δ is, you then proceed with:

    Consider x such that 0 < |x-a| < δ .

    Now you do some algebra, and eventually come up with |x2-a2| < ε.

    * * * * * * * * * * * * * * * * * * * * *
    How do you come up with an expression for δ ?

    As you show in your proof, |x2-a2| = |x-a|∙|x+a|

    The goal here is to have a "bound" on |x+a| . Suppose we can show that |x+a| < B, where B is some number. Then we would have
    |x-a| < δ​
    and
    |x+a| < B​
    So that
    |x-a|∙|x+a| < B∙δ​
    Therefore, if we let δ = ε/B, we have |x-a|∙|x+a| < B∙(ε/B) = ε .

    How do we find that bound on |x+a| ?

    We already know that we will require |x-a| < δ. do some algebra on this:
    -δ < x-a < δ

    2a - δ < x+a < 2a + δ          (Added 2a to each part)

    -(2|a| + δ) ≤ 2a - δ < x+a < 2a + δ ≤ 2|a| + δ

    |x+a| < 2|a| + δ ​

    So you can choose δ so that δ(2|a| + δ) = ε . Solve this for δ, remembering that δ must be positive.

    If you don't like solving that quadratic equation for δ, you could put some restriction on δ, such as δ ≤ 1.

    Then you would know that |x-a| ≤ 1 . Following the same algebraic steps, that gives |x+a| ≤ 2|a| + 1. That would result in δ(2|a| + 1) = ε, which is easily solved for δ . Use the "min" function to make sure that δ ≤ 1 .
     
  5. Jan 4, 2012 #4

    Mark44

    Staff: Mentor

    No, please start a new thread for each problem.
     
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