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Concerning the definition of limit

  1. May 1, 2009 #1
    "In the first place, let me say that delta depends upon epsilon and not the other way round as you have stated. We select an arbitrary epsilon greater than zero and if we succeed in finding delta greater than zero satisfying the dfinition of the limit, then L is the limit. We have to express delta in terms of epsilon and taking the condition that epsilon is positive, prove that delta is positive for every positive epsilon. If, on the other hand, the relation between delta and epsilon so turns out that delta is not positiver for "every" positive epsilon, then we can conclude that L is not the limit. Remember that the definition has to be satisfied for every possible positive epsilon and not just one arbitrary positive epsilon."

    reference:
    http://answers.yahoo.com/question/i...MhyZTxIazKIX;_ylv=3?qid=20090429004411AAa2BJz


    So, I`ve asked about how do we know (what is the proof) that Epsilon is a function of Delta?

    and now I wanna ask the same question here..

    but also, I have an additional question, why is the MR talking about "positive-negative"
    I thought that the limit doesn`t exist when we can`t even make a relation between the two parts of the definition??


    Thank you,
     
    Last edited: May 1, 2009
  2. jcsd
  3. May 2, 2009 #2
    The answer lies in how a mathematical statement is interpreted. In this case, how is a universal quantifier followed by an existential quantifier interpreted? For example, “For every epsilon, there exists a delta…”


    The following is the DEFINITION of how these are interpreted:
    For this example I will use P(x,y) to represent some statement about x and y such as “x<y” or “x is the brother of y”. The exact statement is not relevant to the discussion.

    Consider: "For all x, there exists a y, such that P(x,y)"

    To show this statement is true, you need to let someone pick any x they wish. Then based upon that x, you need to find a y such that P(x,y) is true. Then, they get to go again! They can pick any x, you again have to find a y such that P(x,y) is true. This game goes on until the person has gone through all possible x’s and in each case you found a corresponding y. Is this sense, y “depends” on x. If you succeeded EVERY time an x was chosen (and did so for all x’s in the set) then the statement is true. Note, if for some x they pick, you cannot find a y, then the statement is false. Also, note, if the set of x’s is infinite in size then going thru one by one is ridiculous to do, so another proof strategy is needed. The usual technique is to say “let x be arbitrary and fixed” then based on that “random” x, go find a y that will work. This is why it is necessary to write y as a function of the x. Think of it as instructions that tell the person how to go find the y based upon whatever x they chose.


    Alternately, compare “For all x, there exists a y, such that P(x,y)”
    with the following statement:

    “There exists a y, for all x, P(x,y).”

    Now, for this to be true, you must find some y (fixed!) that will work for every x!!!
     
  4. May 2, 2009 #3

    HallsofIvy

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    The definition of "limit of f(x) as x approaches a" is

    "[itex]\lim_{x\rightarrow a} f(x)= L[/itex] if and only if, for every [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if 0< [itex]|x-a|< \delta[/itex], then [itex]|f(x)- L|< \epsilon[/itex]."

    The "for every [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex]" says that given any [itex]\epsilon[/itex], we can find [itex]\delta[/itex]. That is the part that says [itex]\delta[/itex] depends on [itex]\epsilon[/itex]. It does NOT mean, nor does any part of what you give above, imply that [itex]\delta[/itex] is a function of of [itex]\epsilon[/itex]. The same [itex]\delta[/itex] may apply for many different [itex]\epsilon[/itex].

    Also, that definition requires that "[itex]\epsilon> 0[/itex]" and "[itex]\delta> 0[/itex]" which is why it is talking about "positive". If, given [itex]\epsilon> 0[/itex], you can find a corresponding [itex]\delta[/itex], but it is negative, that's not good enough.
     
  5. May 2, 2009 #4

    matt grime

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    Epsilon, as Halls states (though I believe he meant to say that different delta may work for the same epsilon; functions cannot be one to many, but may be many to one), is not (necessarily) a function of delta in the mathematical sense of the word, but it can be viewed as such in the looser use of the word that exists in everyday English, and in fact one can often see the usefully suggestive notation d(e) to remind the reader that delta depends on epsilon, although that dependency may be trivial, and not actually a function.

    And often the way one proves something is continuous is to find a particular function d(e) in the proper mathematical sense. For example, to show that f(x)=x is continuous then using the function d(e)=e suffices. In fact it is clear that one can actually require that delta be a function of epsilon (just choose a delta for each epsilon to make a function).
     
  6. May 2, 2009 #5

    HallsofIvy

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    Yes, thanks, Matt. Don't know where my mind was.
     
  7. May 3, 2009 #6
     
  8. May 3, 2009 #7

    matt grime

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    f(x)=0 for x=/=0 and f(0)=1 is not continuous at 0 - discontinuities can be thought of as 'steps' in a simplistic way.
     
  9. May 4, 2009 #8
    1) sorry, but in terms of the definition of limit, what are "all" the cases where the limit doesn`t exist? (what are the possibilities?)

    2)
     
  10. May 4, 2009 #9

    matt grime

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    A function that is discontinuous at every point? These are sort of pathological functions, but f(x)=0 when x is rational and 1 when x is irrational is the typical example - it even has a name, but I can't remember it for sure (dirichlet function maybe?).
     
  11. May 6, 2009 #10
    ummm, lemme ask my question in a better way, but I want to know an answer to this question first, plz



    How can it be negative???
     
  12. May 7, 2009 #11
    If you think you have a rule for finding delta given a value for epsilon, then the rule must not cause you to choose a negative value for delta.

    For example: delta = epsilon – 1

    This would never work for a rule in the limit argument because
    when we pick epsilon = .5 we get delta = -.5. Not allowed.

    The definition of limit requires both epsilon and delta to be positive (think of them as distances.)

    As far as a function not having a limit at a certain point, there are several ways this can happen.

    When f(x) does not have a limit at x = a it could be due to one of these cases:

    1) f(x) is not even defined on an interval around a.
    For example, ln(x) is not defined for x < 0 (nor x=0).
    So, ln(x) cannot have a limit at x = -1.

    2) f(x) is unbounded as it approaches x=a.
    For example, f(x)=1/x does not have a limit at x=0. This is because, f(x) climbs forever upwards from the right side of 0. Also, it falls forever downward from the left side of 0. Either argument would indicate that 1/x has no limit at 0.

    3) f(x) keeps “jumping” around as you get close to x = a. This is a messier situation to describe. It happens when there are two values (at least two actually) that f(x) tries to go to as x approaches a. Say, f(x) tries to go to 1 and it tries to go to 0 as x approaches a. So, there are infinitely many points on the graph that are close to height 1 and infinitely many that are close to the height 0 no matter what interval around x=a you are in. This is like the example mentioned earlier in the post (f(x) = 1 if x is irrational and f(x) = 0 when x is rational.)
    Another example is f(x) = sin(1/x). In this case, as you get close to x = 0, f(x) will continually jump from -1 to 1. No matter how small the interval you choose around 0, say (-1/N , 1/N), f(x) will be at height 1 and at height -1 an infinite amount of times. So, f(x) is not limiting to any specific value as it approaches 0.
     
  13. May 8, 2009 #12
    Well, if this is the definition of Limit:
    for each real ε > 0 there exists a real δ > 0 such that for all x with 0 < |x − c| < δ, we have |f(x) − L| < ε.

    Then how would I ever get a negative delta..
    on the right sides we want delta and epsilon to be greater than 0, so epsilon is already positive.
    on the left sides, we have absolute values so we can`t get any negative value.
    well, the only possibilty I can think of is that we may need to solve |f(x) − L| for some cases and put it on the form [f(x) − L]&-[f(x) − L]
    So, can you give me a function and the flow of the proof where we will end with haveing a negative value?
     
  14. May 8, 2009 #13

    HallsofIvy

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    You wouldn't. If you did, then you made a mistake. That was my point!

    To show that [itex]x^2[/itex] is continous at 0, you must show that, given some [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]|x|< \delta[/itex], then [itex]|x^2|< \epsilon[/itex]. You might find [itex]\delta[/itex] by taking the square root of both sides: [itex]x< \pm \sqrt{\epsilon}[/itex]. If you took [itex]\delta= -\sqrt{\epsilon}[/itex] you would have a negative value for [itex]\delta[/itex]. That would, of course, be a mistake.
     
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