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Epsilon Delta Definition of Limit

  1. May 31, 2012 #1
    So far, all I understand is that the definition proves that there's a value of f(x,y) as f(x,y) approaches (x0,y0) which is sufficiently close to but not exactly the value at f(x0,y0). I am probably completely off... but I just don't understand the purpose of proving this. I also don't understand the purpose of the ε and the δ... which is pretty much the whole definition. Please help! D:
     
  2. jcsd
  3. May 31, 2012 #2

    HallsofIvy

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    No, "sufficiently close" is too vague. [itex]\lim_{x\to a}f(x)= L[/itex] means that we can get the value of f(x) "arbitrarily close" to L by getting x close enough to a. We can think of |a- b| as "the distance between a and b" so that "[itex]|x- a|< \delta[/itex]" says "x is closer to a than distance [itex]\delta[/itex]" and "[itex]|f(x)- L|< \epsilon[/itex]" says "f(x) is closer to L than distance [itex]\epsilon[/itex]".
     
  4. May 31, 2012 #3
    I still can't see how this is the definition of a limit.
     
  5. May 31, 2012 #4
    If you think of a function f which goes from [itex]\Re[/itex] to [itex]\Re[/itex] that definition means that a function has a limit in one point (x) if if you get close enough to that point (d(t,x)<δ) then the image of that point by that function will be as close as you want to the limit (d(f(a),L)<ε).
     
  6. May 31, 2012 #5
    I had a tedious time understand the precise definition of a limit too. Here is how I had to think about it; maybe it will work for you.

    Limx→a f(x) = L
    We know that the limit of a function at a is the value f(x) approaches as x approaches a. To prove this mathematically we employ a few tactics.

    First you must understand that there is a relationship between input and output values of a function defined by the rule of a function. (I’m sure you get this part).

    Next you need to realize that there is a relationship in the distance between inputs of a function and the distance between outputs of a function. Here is where ε and δ come into play.

    We utilize two absolute value inequalities to express the relationship between the distance of input values and the distance of output values.

    Here is an example.

    Given a function:

    f(x)=2x+1

    We want to know the limit as x→2.

    We determine that

    Limx→2f(x)=5

    Now we must prove that no matter how close we want to get to 5 we can get that close by making x closer to a.

    Enter ε and δ.

    ε represents the cap on the distance between f(x) and L (outputs).

    δ represents the cap on the distance between x and a (inputs).

    It’s important to understand that we want ε to determine δ. We want to show that no matter how small we make ε there is a positive real number we can make δ to make the following statement true.

    If |x-a|< δ then |f(x)-L|< ε

    here we are saying that if the distance between x and a is less than some arbitrary number δ then the distance between f(x) and L will be less than some other arbitrary number ε.

    Our goal from here is to understand the relationship between ε and δ for this function.

    We do that in the following way.

    If |x-2|< δ then |(2x+1)-5|< ε

    Again we must determine the relationship of the distance between inputs and the distance between outputs.

    We do this by using the rules of absolute values and simplifying:

    |2x+1-5| = |2x-4| = |2||x-2|< ε We can rewrite our previous statement:

    If |x-2|< δ then |2||x-2| < ε

    Here we used factoring to show that, for this point on the function f(x), the distance between outputs is twice the distance between inputs.

    From this you can tell that the distance between inputs is ½ the distance between outputs by doing the algebra and dividing both sides by 2.

    |x-2|< [itex]\frac{ε}{2}[/itex]

    Here we have shown that the distance between inputs is always ½ the distance between outputs.

    So from this it looks like we should choose:

    δ=[itex]\frac{ε}{2}[/itex]

    but we MUST prove this.

    We prove this by showing that we can go from:

    |x-2|< δ

    to

    |(2x+1)-5|< ε

    |x-2|< δ= [itex]\frac{ε}{2}[/itex] multiply both sides by 2
    2|x-2|< ε Distribute the 2
    |2x-4|< ε change the -4 into the form we know from the original function
    |2x+1-5|< ε

    By using ε and δ as variables instead of assigning them values we have shown that we can make ε (the cap on the distance between f(x) and L) as small as we like (any real number) and there will still be a real number δ (which equals [itex]\frac{ε}{2}[/itex]) which will represent the cap on how close we must get x to a. The logic of this proof is that you must realize that any real number divided by a real number (except 0) is a real number.

    For me, though, the proof was not so important as understand that we are determining the relationship of the distance between inputs and the distance between outputs, and the fact that there was a relationship between the distance of inputs and distance of outputs to begin with.

    Play with the math and play with some other linear function to get a feel for this relationship. Moving on to function like f(x) = x2 is the same as with linear functions, but more steps are required. However if you understand how to determine the relationship we've talked about and what that relationship means you will have no problem.

    Hope that helps. Sorry I couldn't explain it how I understand it in fewer words.
     
  7. May 31, 2012 #6
    It's handy to have the definition in front of you when talking about it, so here it is:

    The intuition behind limits and continuity is actually pretty interesting. It models our understanding of scientific measurements. Let me explain.

    In science, the accuracy of our results are always limited by how good our tools are. If you want to measure the distance between two points, you can use your feet ("it's 30 footsteps to the kitchen"), you can use a ruler ("it's 25 feet and 2 inches to the kitchen") or you can use some more accurate tool ("it's 7,670,560 microns to the kitchen"). It's understood no tool gives you perfect accuracy, and there's always a tolerance. ("The distance to the kitchen is between 25 feet and 25 feet 6 inches to the kitchen").

    If you want a more precise measurement, you need a more precise tool. And that's essentially what ε and δ represent in the equation.

    ε is how accurate you want your outcome.

    δ is how accurate your device needs to be to achieve an accuracy of ε.

    Both ε and δ are required to be nonzero. This corresponds to the assumption that no measurement is ever perfect.

    The absolute values are also a kind of short hand. In the last part, "|f(x) - L| < ε" means that the measurement you end up taking, f(x) is between L + ε and f(c) - ε, where L is the "actual value" of the value you're trying to measure.

    The part "if |x - c| < δ" is a conditional statement. It role in the analogy is that you did indeed take a measurement with a sufficiently accurate tool.

    So that's the intuition.

    Logically speaking, ε-δ proofs are as tricky as you can get. It's a tricky mix of universal and existential quantifiers and logical implication (the if |x - c| < δ part) which is often given the "wrong" way. The order of quantifiers and the direction of implications is critical to get right, and the usual quasi-natural language does a garbagety job to communicate it.

    A few tips on doing ε-δ proofs yourself:

    Your goal is to produce a value of δ. δ is actually just an expression defined in terms of f, c, L, and ε (the variables that are in scope where δ is "created").

    Once you've settled on a value for δ, the facts ε > 0 and |x - c| < δ come "for free"... your only job is to use those two facts to show that |f(x) - L| < ε.

    If you can do those two things (find a value for δ in terms of f, c, L, and ε and prove |x - c| < δ), you have shown c is, in fact, the limit of f approaching L.
     
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