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Limit : alternate defination

  1. Nov 27, 2012 #1

    ato

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    limit : alternate defination

    1. this is my attempt to redefine limits in such a way that it remove's following problems. for only defination skip to point 7.

    2. limit textbook defination : assuming f(c) is not defined . lim (f(x)) at c is deduced by taking value of x as close to c as possible but not c.

    3. the problem 1: 'as close to as possible', what does that even mean ? how much close is possible ? how do you determine that ?
    is x + ((x-c)/2) close enough ? the inclusion of the term 'possible' without defining what does it mean mathematically
    is annoying . why would even we used used the word 'possible' ? it seems like the statement ask how much one is capable of getting
    x close ?
    so overall i hate 'possible' and the aim of this post is to remove that 'possible' .

    4. the problem 2: since i dont know which when is closest, i dont know when i am wrong . since the only way to prove i am right
    is to prove that i cannot be wrong, so i dont know if i am right.
    lets say f(c) = 0/0 but lim (f(c)) = L , then according to current limit defination
    nothing => [ [ b =/= L] => [ lim f(c) =/= b ] ]

    5. there are variables (x), when they change from one value (x_1) to another value (x_2), they go through all the value between x_1 and x_2. they
    dont skip a value. for example speed (s) of a ball when increases from s_1 to s_2 then s takes all values in real subset [s_1,s_2].
    lets call such variable natural_variable .

    6. lets define a variable y = ((x^2) - (a^2))/(x-a) AND domain_x = R .
    but turns out y(a) = 0/0 . however it is not a problem or contradiction or paradox . but you cant say [ y is natural_variable ] which is a
    problem because natural_variables are to be studied . or in other word y has to be a natural_variable or to be made into one.
    so to solve this we need to make sure,
    i. nothing implies f(a) = 0/0
    ii. assign a value to f(a) , otherwise it would still be a non natural_variable.

    7. limit, alternate defination :
    assign lim (f(x)) at c in such a way that
    for every x
    f takes all value between f(x) and ( lim f(x) at c ) within x and c.

    8. limit, alternate def - mathematical version :
    [ lim f(x) at c = L ] <=> [ for all x in domain_x, domain_y_x n R = domain_y_x, where domain_y_x is set of all values of f(x) between x and c ]

    NOTE:
    i. n is set intersection sign
    ii. examples of domain_y_x
    domain_y_1 for c = 2:
    x_i belongs to [1,2] => (domain_y_1 for c equals 2) = {f(x_1), f(x_2), ... , f(x_n-1), L}
    domain_y_1 for c = 0
    x_i belongs to [0,1] => (domain_y_1 for c equals 0) = {f(x_1), f(x_2), ... , f(x_n-1), L}

    9. so new_lim proves why we cannot assign lim f(x) at a = 1 if f(x) = ((x^2) - (a^2))/(x-a) . because f(x) is not equal to 1/2 between
    0 and 1 -a but should have because f(0) > 1/2 > f(1-a).

    so here it is. the redefination serves its purpose. i hope it helps anyone having those problems.

    any comment/contradiction is welcome.
     
  2. jcsd
  3. Nov 27, 2012 #2

    pwsnafu

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    Welcome to Physics Forums. I recommend you install a spell checker in your browser.

    You claim to redefine the concept of "limit of a function". I implore you to reread (and understand) the actual definition first, because none of your criticisms from points 3 & 4 apply.
     
  4. Nov 28, 2012 #3

    HallsofIvy

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    This is not true. Whether f(c) is defined or not, and, if it is, what its value is, is not relevant to "the limit as x goes to c".

    No textbook I have ever seen uses the phrase "as close as possible". Some will use the phrase "close to" when describing the limit concept but then make that precise in the actual definition: given some [itex]\delta> 0[/itex], [itex]|x- c|< \delta[/itex].

    We don't. You seem to be attacking a straw man. Perhaps you are simply misreading a textbook.

    Then the purpose of this post is to remove something that isn't there to begin with!

    No, we can't. That's impossible. Text books will give examples of f(x)= g(x)/h(x) and consider cases where g(x)-> 0 and h(x)-> 0 but that is NOT saying that "f(c)= 0/0". Again, you are misreading your text book.

    NO!

    Why not use the standard terminology and call it a continuous variable?

    You can't so you are starting with an impossible situation.

    Since your hypothesis is impossible, everything you say here is meaningless.

    What does "within x and c" mean? You are the one being vague here.

    I have no idea what you mean here but you seem to be defining the limit as a set of values of f between x and c. But this is supposed to be the limit as x goes to c so what value of x are you using here?

    I can't make enough grammatica sense out of this statement to comment.
    "but should have because"- "should have" what?

    Unfortunately this is just what we see here regularly: you are saying "I do not understand it, therefore it is wrong". Take a good Calculus course and try understanding!

    (And, by the way, you are consistently mis-spelling "definition".)
     
    Last edited: Nov 28, 2012
  5. Nov 28, 2012 #4

    Fredrik

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    This is not the textbook definition. The standard definition is the one that pwsnafu linked to. I agree that a definition that uses ideas like "as close as possible" without further explanation is completely useless. But I have never seen such a definition in a book.

    Here's a definition that's equivalent to the epsilon-delta definition:

    Let c be any number such that every open interval that contains c also contains a member of the domain of f. f is said to have a limit at c if there's a number y such that for every open interval A that contains y, there's an open interval B that contains c and is such that f(B) is a subset of A. The number y is called "the limit of f(x) as x goes to c" and is denoted by ##\lim_{x\to c} f(x)##.

    Here f(B) denotes the set of all f(x) such that x is in B.
     
  6. Nov 29, 2012 #5

    ato

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    -------------

    here's what i think , you are saying
    [
    [
    for all open interval A
    [ y belongs to A ] =>
    [
    for at least one open interval B
    [ c belongs to B ] AND [b_i belongs to B] AND [ f(B) = {f(b_1),f(b_2), ... , f(b_n)} ] AND [ {f(b_1),f(b_2), ... , f(b_n)} is subset of A ]
    ]
    ] => [ lim f(x) as x goes to c is y ]
    ] ... [1]

    so lets prove [ lim f(x) as x goes to c is y_0 ] assuming [ [ lim f(x) as x goes to c is y ] AND [y =/= y_0] ]

    but statement [1] will help us do that provided we
    prove
    [
    for all open interval A
    =>
    [
    for at least one open interval B
    [ c belongs to B ] AND [b_i belongs to B] AND [ f(B) = {f(b_1),f(b_2), ... , f(b_n)} ] AND [ {f(b_1),f(b_2), ... , f(b_n)} is subset of A ]
    ]
    ]

    -> prove
    [
    for at least one open interval B
    [ c belongs to B ] AND [ {f(b_1),f(b_2), ... , f(b_n)} is subset of A ]
    ]
    assuming [ [ y_0 belongs to A ] AND [b_i belongs to B] AND [ f(B) = {f(b_1),f(b_2), ... , f(b_n)} ] ]

    {f(b_1),f(b_2), ... , f(b_n)} is not an interval because
    [ [{f(b_1),f(b_2), ... , y , ... , f(b_n)} is interval ] AND [ y =/= y+0 ] ] => [ {f(b_1),f(b_2), ... , y_0 , ... , f(b_n)} is not interval ]

    in other words you cant prove [ {f(b_1),f(b_2), ... , y_0 , ... , f(b_n)} i interval ]

    -> so you cant prove the original statement [ lim f(x) as x goes to c is y_0 ].

    this solves :
    the problem 1 because it does not rely on phrases like "as goes to" , "as approaches to" or "as close to as we like" or "as close as possible"
    ( and at least 3 out 4 of them has been used multiple times in , stewart calculus 5e) , and
    the problem 2 , it says why you cant prove otherwise .


    i think its equivalent to what i said . however it be good if i come across this definition earlier.

    -------------

    epsilon_delta limit statement:
    For every real ε > 0, there exists a real δ > 0 such that for all real x, 0 < | x − p | < δ implies | f(x) − L | < ε

    it will still be correct if i replace L with L_wrong.
    For every real ε > 0, there exists a real δ > 0 such that for all real x, 0 < | x − p | < δ implies | f(x) − L_wrong | < ε

    so it does not solve problem 2 mentioned in point 4. in other words, i need an definition/statement/algorithm which would either tell me
    1. why L is correct limit ?
    or/and
    2. why L_wrong is wrong limit ?

    -------------

    yes i know the assumption is not needed. but i was going at the root of the problem because if we did not need to define the undefined we would not have invented limits.
    or in other words, f(c) is not defined means (0/0 , infi/infi etc ), limit is what i said above ("textbook definition") . and if f(c) is defined then limit is what you
    would usually do following usual operations.

    -------------

    "f(c)= 0/0" is result of more fundamental assumptions that i have took. those assumption does not prove false that
    "f(x)= g(x)/h(x) and consider cases where g(x)-> 0 and h(x)-> 0" however/so i can use this too. i think its matter of syntax.

    ------------

    i hope you would give something to support this. otherwise it has no meaning to me.

    ------------

    i dont know if it is equivalent to continuous function defined by using limit ("the textbook version").
    i took the safe route and define it something else. however i am not saying
    [ x is natural_variable ] => [ x is not continuous variable ] but in future i might say
    [x is natural variable ] <=> [ x is continuous variable ] .

    -------------

    how ? but if you have problem with the domain_x , you can just replace it with any other real interval which includes a.

    -------------

    within x and c <=> from a real interval which include x and c as its endpoint.

    -------------

    domain_x is domain of f , is set of variables that we put in f to get y
    domain_y is the set of variables that we get from f after we put every variable in domain_x into f. [ domain_y is real interval ] <=> [ y is natural_variable/continuous_variable ]
    the aim of the limit is to keep the domain_y a real interval.
    domain_y_x is the set of {f(x_1),f(x_2) , ... , f(x_n)} where x_i belongs to the real interval taking x and L as endpoints.

    [
    for all x in domain_x,
    [ domain_y_x n R = domain_y_x ]
    ]
    since A n R = A <=> A is real interval.so above is equivalent to
    [
    for all x in domain_x
    [ domain_y_x is real interval ]
    ]
    equivalent to
    [
    [domain_y_(x_1) is real interval ] AND [domain_y_(x_2) is real interval ] AND ... AND [domain_y_(x_n) is real interval ]
    ] assuming domain_x = {x_1,x_2, ... , x_n}

    so

    [ [ lim f(x) at c = L ] AND [ domain_x = {x_1,x_2, ... , x_n} ] ] <=>
    [ [domain_y_(x_1) is real interval ] AND [domain_y_(x_2) is real interval ] AND ... AND [domain_y_(x_n) is real interval ] ]

    so if you could prove this [ [domain_y_(x_1) is real interval ] AND [domain_y_(x_2) is real interval ] AND ... AND [domain_y_(x_n) is real interval ] ]
    then if you could prove for a L then L is limit.

    --------------

    i apologies point 9 is incomplete and wrong , correct version,

    9. So new_lim proves why we cannot assign lim f(x) at a if f(x) = ((x^2) - (a^2))/(x-a) to a/2. because then we cant prove [ domain_y_0 is a real interval ]

    -------------

    i also was not going for the literal syntax. some phrases like
    "as close to we like "
    "as close to a number as we like "
    "as close to as we please"
    "as approaches to"
    "taking a number sufficiently close"
    "a->b"
    "by taking a small number" # there
    "by taking a large number" #
    that i know right that is at least used in books like stewart 5e, wikipedia , marsden calculus 1e . there reference is not just limited to calculus.
    the real problem any statement that have these phrases are 'useless' statements, they prove anything true or false.

    ------------

    i am not redefining anything such as it would break all assumptions of calculus.
    nor am i trying to start a new calculus like a religion .
    i am just trying to deduce a equivalent statement that solves the problems , since i think have, i am here for a peer review .

    -------------

    i did not expect it to be a problem. anyway i have installed a spell checker now.

    -------------
     
  7. Nov 29, 2012 #6

    Fredrik

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    There are at least two problems with your statement:
    1. When you mention b_i the first time, you haven't defined it.
    2. f(B) is usually not finite, so the notation f(B) = {f(b_1),f(b_2), ... , f(b_n)} is inappropriate. You can however write f(B)={f(x)|x is in B}.

    I don't think your computer code style makes it easier to understand the statement, but I'll show you a way to rewrite it.
    [
    [
    for all open interval A
    [ y belongs to A ] =>
    [
    for at least one open interval B
    [ c belongs to B ] AND [for all b that belongs to B [f(b) belongs to A]]
    ]
    ] <=> [ lim f(x) as x goes to c is y ]
    ] ... [1]

    It looks like your point is that this can not be done. In that case you're right. My definition (as well as the equivalent epsilon-delta definition) does ensure that limits are unique. This is one way to prove it:

    Suppose (to obtain a contradiction) that f(x) has two limits y and z (with y≠z) as x goes to c. Let A be an arbitrary interval that contains y but not z. (Such an interval exists, since we can define ε=|z-y|/2, and take A=(y-ε,y+ε)). Let B be an interval that contains c and is such that f(B) is a subset of A. (Such an interval must exist since we have assumed that f(x)→y as x→c). Now let A' be an arbitrary interval that contains z and is disjoint from A. Now there is clearly no interval B' that contains c and is such that f(B') is a subset of A', because every interval B' that contains c will also contain members of B, and f takes them to members of f(B), which is a subset of A, which is disjoint from A'. This contradicts the assumption that f(x)→z as x→c.

    I recommend that you try to prove that the epsilon-delta definition also ensures that limits are unique. Since that definition is very similar to mine, the proof will be very similar to mine.

    Textbook definitions do not rely on phrases like "goes to" or "approaches". In fact, those definitions are what assigns a meaning to those phrases.
     
  8. Nov 29, 2012 #7

    micromass

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    I suggest you get a better textbook than the awful book by Stewart. Try to get a hold of Spivak or Apostol. They give the actual definition of continuity and limits.

    That's not what this forum is for. We can try to explain limits and epsilon-delta to you. But if you want to discuss original and personal ideas, then you can't do it here.
     
  9. Nov 29, 2012 #8

    pwsnafu

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    1. L is the correct limit if it satisfies the property that you quoted.
    2. We can prove: if Lwrong also satisfies the property, then L = Lwrong.

    Edit:

    Sorry, but I don't believe you. Your original post made absolutely no attempt to prove that your definition was equivalent to epsilon-delta.
     
    Last edited: Nov 29, 2012
  10. Nov 29, 2012 #9
    I don't really want to defend Stewart, but he does have the correct definitions of limits and derivatives. I have plenty of issues with his books too, but I don't think he is the problem here.
     
  11. Nov 29, 2012 #10

    Mark44

    Staff: Mentor

    0/0 is undefined, so you cannot say as you did, that f(c) = 0/0. Period.

    If this is the result of "more fundamental assumptions" then those assumptions are not valid and should be thrown out.



    HallsOfIvy did support what he said; namely that you cannot write f(c) = 0/0 and think that it means something.

    Further, you have something pretty incomprehensible; namely
    I have no idea what you're trying to say with that. It is logically meaningless.
     
  12. Nov 30, 2012 #11

    ato

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    when i read for this version of epsilon delta definition
    this is what i see and understand (please tell me if it is incorrect),
    [
    for all x
    [ δi is the ith element of set (0,∞) ] AND [ εi is the ith element of set (0,∞) ] AND
    [ [ 0 < |x - a| < δ1 ⇒ |f(x) - L| < ε1 ] OR [ 0 < |x - a| < δ2 ⇒ |f(x) - L| < ε1 ] OR ... OR [ 0 < |x - a| < δ ⇒ |f(x) - L| < ε1 ] ] AND
    [ [ 0 < |x - a| < δ1 ⇒ |f(x) - L| < ε2 ] OR [ 0 < |x - a| < δ2 ⇒ |f(x) - L| < ε2 ] OR ... OR [ 0 < |x - a| < δ ⇒ |f(x) - L| < ε2 ] ] AND
    ...
    [ [ 0 < |x - a| < δ1 ⇒ |f(x) - L| < ε ] OR [ 0 < |x - a| < δ2 ⇒ |f(x) - L| < ε ] OR ... OR [ 0 < |x - a| < δ ⇒ |f(x) - L| < ε ] ]
    ] => [ limx->af(x)=L]

    and i am sure ( still i have to prove though but first please tell me if this form is actually the epsilon delta limit ) that
    it proves L1 and L2 is limit for same funtion at same number, even if L1 ≠ L2.

    however if i read this
    i see either (because i am not sure ε band means exactly what between f([f-1(L-ε),f-1(L+ε)]) and [L-ε,L+ε])
    [
    for all ε < 0, for at least one δ < 0
    f([a-δ,a+δ]) ⊂ [L-ε,L+ε]
    ]
    or
    [
    for all ε < 0, for at least one δ < 0
    f([a-δ,a+δ]) ⊂ f([f-1(L-ε),f-1(L+ε)])
    ] assuming f-1(L-ε) and f-1(L-ε) is unique

    but since
    [
    for all ε < 0, for at least one δ < 0
    f([a-δ,a+δ]) ⊂ f([f-1(L-ε),f-1L+ε])
    ]
    is always true irrespective of L which means it can prove two different limit for same function at same number.
    so i am going with
    [
    for all ε < 0, for at least one δ < 0
    f([a-δ,a+δ]) ⊂ [L-ε,L+ε]
    ]

    which basically says f([a-δ,a+δ]) has to be a real interval otherwise limit is wrong which is what i said,
    "lim f(x) at a can be anything you have to make sure that f range remains/becomes a real interval" or more formally
    [
    f(domainf) ⊂ R
    ] assuming f has only one hole (undefined value)
    and
    [
    for at least one δ < 0
    f([a-δ,a+δ]) ⊂ R
    ] assuming f has any number of hole

    and i think its better and more easy to translate in to what we are trying to achieve.

    1. b1 is 1st element of B and b2 is 2nd element of B and so on .
    2. i mean , f(B)={f(x) | x is in B }

    what do you mean by
    f takes them to members of f(B) ?

    because i never said that the deduction is equivalent to epsilon-delta. howevery the deduction is equivalent to
    [ lim x/x at 0 is 1 and lim (x^2-x^1)/(x-1) at 1 is 2 and ... ] . because i already said i am not trying to break down the whole calculus .

    but if you believe [the deduction ⇔ [ lim x/x at 0 is 1 and lim (x^2-x^1)/(x-1) at 1 is 2 and ... ] ] and
    [ epsilon-delta ⇔[ lim x/x at 0 is 1 and lim (x^2-x^1)/(x-1) at 1 is 2 and ... ] ] and
    [ [ [ statementA ⇔ statementB] AND [ statementA ⇔ statementC ] ] ⇒ [ statementB ⇔ statementC] ]
    then
    [ the deduction ⇔ epsilon delta ] .


    i already said by f(c) = 0/0 i mean only and only f(c) goes to 0/0 as x goes to c.
    or better forget i said f(c) = 0/0 , instead i say f(c) is undefined and thats is only what i meant actually as you can see i have not used f(c) = 0/0 anywhere else .

    what i meant is i cannot find a statement that i could replace with "nothing" such as [ replacing_statement => [ [ b =/= L] => [ lim f(c) =/= b ]] ] is true.
    what i meant by [ [ b =/= L] => [ lim f(c) =/= b ]] is [ [ [ b =/= L ] and [ lim f(x) at c = L] ] => [ lim f(x) at c =/= b ]] however this is always true. which is why you might thought
    is meaningless. however what i actually wanted to say is
    [ replacing_statement => lim f(x) at c = L ] and [ L =/= Lwrong ] and [ replacing_statement =/> lim f(x) at c = Lwrong ]
     
  13. Nov 30, 2012 #12

    Fredrik

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    I'm going to stop you right there, because the members of the set (0,∞) can't be labeled by integers. (0,∞) is not countable.

    This idea doesn't work, because there's no way to assign a unique positive integer to each member of B. In other words, B is not countable. Because of this, you have to change the way you talk about these sets.

    The set of rational numbers is countable, but the set of real numbers is not. An interval of real numbers is not countable.

    I'm just saying that for all x in ##B'\cap B##, we have ##f(x)\in f(B)##. This follows immediately from the definition of f(B) and the fact that ##B'\cap B\subset B##.

    In general, if f is a function and x is a member of the domain of f, then f(x) is called the value of f at x, and it's common to say that f takes x to f(x). That's how I use the word "takes". For example, the function ##f:\mathbb R\to\mathbb R## defined by ##f(x)=x^2## can be described as "the function that takes every real number to its square". In particular, it takes 3 to 9.

    Edit: It's probably just confusing for you to have to look at two different definitions, so I suggest that you focus on the epsilon-delta definition for now. If you want to discuss my definition or other equivalent definitions, I suggest that we do it after you have understood the epsilon-delta definition.
     
    Last edited: Nov 30, 2012
  14. Nov 30, 2012 #13

    Fredrik

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    Since you really seem to want to see the definition of "f(x)→L as x→a" in computer code style...

    Code (Text):

    For all ε
    [
      if ε>0 then there's a δ such that
      [
        δ>0
      and
        for all x [ if 0 < |x − a | < δ then |f(x)-L| < ε ]
      ]
    ]
     
     
    Last edited: Nov 30, 2012
  15. Nov 30, 2012 #14

    pwsnafu

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    RHS does not imply LHS. All you are doing is looking at the removable singularities of rational functions.
     
  16. Nov 30, 2012 #15

    ato

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    please dont. read it all first if you have not. because what i am going to say there does not depend upon countability. or else you could show me what contradictions would it bring, for example
    in this statement
    [ [ 0 < |x - a| < δ1 ⇒ |f(x) - L| < ε1 ] OR [ 0 < |x - a| < δ2 ⇒ |f(x) - L| < ε1 ] OR ... OR [ 0 < |x - a| < δ ⇒ |f(x) - L| < ε1 ] ]

    or even more better ( hoping that it would priotize your response better ) ignore this whole statement in the previous post
    [
    for all x
    [ δi is the ith element of set (0,∞) ] AND [ εi is the ith element of set (0,∞) ] AND
    [ [ 0 < |x - a| < δ1 ⇒ |f(x) - L| < ε1 ] OR [ 0 < |x - a| < δ2 ⇒ |f(x) - L| < ε1 ] OR ... OR [ 0 < |x - a| < δ ⇒ |f(x) - L| < ε1 ] ] AND
    [ [ 0 < |x - a| < δ1 ⇒ |f(x) - L| < ε2 ] OR [ 0 < |x - a| < δ2 ⇒ |f(x) - L| < ε2 ] OR ... OR [ 0 < |x - a| < δ ⇒ |f(x) - L| < ε2 ] ] AND
    ...
    [ [ 0 < |x - a| < δ1 ⇒ |f(x) - L| < ε ] OR [ 0 < |x - a| < δ2 ⇒ |f(x) - L| < ε ] OR ... OR [ 0 < |x - a| < δ ⇒ |f(x) - L| < ε ] ]
    ] => [ limx->af(x)=L]

    i really dont know you mean by "computer style code" .
    1. using AND,OR comes under logic. it helps to say what we want say to in more flexible form .
    2. using square brackets are needed to show exactly which statements are joined by AND OR => because A => B AND C
    could either be misunderstood as
    A => [B AND C]
    or
    [ A => B ] AND C

    i think there is a communication problem between us .
    i never said [ epsilon-delta ⇔[ lim x/x at 0 is 1 and lim (x^2-x^1)/(x-1) at 1 is 2 and ... ] ] is true.
    i said if you think it is true.
     
  17. Nov 30, 2012 #16

    ato

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    i dont want to now because what ever it can say, following statement says too and i just need a confirmation .
    and i think its better and more easy to translate in to what we are trying to achieve i.e.
    to generalize following statement
    [ lim x/x at 0 is 1 and lim (x^2-x^1)/(x-1) at 1 is 2 and ... ]

    [IGNORE] please ignore this in my previous post its not what i wanted to say.
    [
    f(domainf) ⊂ R
    ] assuming f has only one hole (undefined value)
    and
    [
    for at least one δ > 0
    f([a-δ,a+δ]) ⊂ R
    ] assuming f has any number of hole
    [/IGNORE]
     
  18. Nov 30, 2012 #17

    Fredrik

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    I have no idea what you mean by δ1, ε1, δ2, and so on.
     
  19. Nov 30, 2012 #18

    pwsnafu

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    You are writing in a format similar to the pseudo-code used by programmers. If you want epsilon-delta in symbolic logic, it'll look more like the following:
    ##(\forall \epsilon \, : \epsilon > 0)(\exists \delta \, : \, \delta > 0)(\forall x)((0 < |x-a| < \delta)\rightarrow(0<|f(x)-L|<\epsilon))##


    I'm pretty sure Fredrik knows what conjunction and disjunction are.

    We use round brackets actually.

    The first one. Conjunction has higher precedence than implication.
     
  20. Nov 30, 2012 #19

    ato

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    i mean δ1 means first element of (0,∞) , but i think you are going to ask δ1
    or δ2 is. i would say i dont know , i dont need to know. what i know is (0,∞) is set and
    if it has at least two element then δ1 and δ2 exist and using them in a statement is not a problem.

    however i would insist/request you/anyone to give confirmation on
    update:
    it is wrong because for example
    f([0-π,0+π]) is a real interval even if lim f(x) at 0 = 1/2 where f(x) = sin x .
     
  21. Nov 30, 2012 #20

    Fredrik

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    But you're not just talking about two deltas, you're talking about an infinite sequence of deltas. What is the significance of that sequence? If you think that there's an infinite sequence ##\delta_1,\delta_2,\dots## such that for all x in (0,∞) there's a positive integer n such that ##\delta_n=x##, then you're wrong. That's what "(0,∞) is not countable" means.

    If you meant that ##\delta_1,\delta_2,\dots## is just some arbitrary sequence in (0,∞), then you need to say so. However, I don't see a reason to bring a sequence into this. I thought you were just trying to rewrite what the epsilon-delta definition is saying in a way that you're more comfortable with. And the epsilon-delta definition doesn't mention any sequences.

    The statement to the right of "if and only if" is not equivalent to ##\lim_{x\to a}f(x)=L##. The only observation I needed to make to know that for sure is that it doesn't contain a or L.
     
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