
#1
Nov2712, 09:55 PM

P: 30

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 + ((xc)/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))/(xa) 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_n1), 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_n1), L} 9. so new_lim proves why we cannot assign lim f(x) at a = 1 if f(x) = ((x^2)  (a^2))/(xa) . because f(x) is not equal to 1/2 between 0 and 1 a but should have because f(0) > 1/2 > f(1a). so here it is. the redefination serves its purpose. i hope it helps anyone having those problems. any comment/contradiction is welcome. 



#2
Nov2712, 10:12 PM

Sci Advisor
P: 779

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. 



#3
Nov2812, 07:42 AM

Math
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PF Gold
P: 38,894

"but should have because" "should have" what? (And, by the way, you are consistently misspelling "definition".) 



#4
Nov2812, 08:28 AM

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PF Gold
P: 9,010

limit : alternate definationHere's a definition that's equivalent to the epsilondelta 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. 



#5
Nov2912, 12:09 AM

P: 30


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



#6
Nov2912, 12:49 AM

Emeritus
Sci Advisor
PF Gold
P: 9,010

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] 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=zy/2, and take A=(yε,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 epsilondelta definition also ensures that limits are unique. Since that definition is very similar to mine, the proof will be very similar to mine. 



#7
Nov2912, 02:08 AM

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P: 16,690





#8
Nov2912, 02:16 AM

Sci Advisor
P: 779

2. We can prove: if L_{wrong} also satisfies the property, then L = L_{wrong}. Edit: 



#9
Nov2912, 05:29 PM

P: 427





#10
Nov2912, 06:24 PM

Mentor
P: 21,059

If this is the result of "more fundamental assumptions" then those assumptions are not valid and should be thrown out. Further, you have something pretty incomprehensible; namely 



#11
Nov3012, 07:24 AM

P: 30

when i read for this version of epsilon delta definition
[ for all x [ δ_{i} is the ith element of set (0,∞) ] AND [ ε_{i} is the ith element of set (0,∞) ] AND [ [ 0 < x  a < δ_{1} ⇒ f(x)  L < ε_{1} ] OR [ 0 < x  a < δ_{2} ⇒ f(x)  L < ε_{1} ] OR ... OR [ 0 < x  a < δ_{∞} ⇒ f(x)  L < ε_{1} ] ] AND [ [ 0 < x  a < δ_{1} ⇒ f(x)  L < ε_{2} ] OR [ 0 < x  a < δ_{2} ⇒ f(x)  L < ε_{2} ] OR ... OR [ 0 < x  a < δ_{∞} ⇒ f(x)  L < ε_{2} ] ] AND ... [ [ 0 < x  a < δ_{1} ⇒ f(x)  L < ε_{∞} ] OR [ 0 < x  a < δ_{2} ⇒ f(x)  L < ε_{∞} ] OR ... OR [ 0 < x  a < δ_{∞} ⇒ f(x)  L < ε_{∞} ] ] ] => [ lim_{x>a}f(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 L_{1} and L_{2} is limit for same funtion at same number, even if L_{1} ≠ L_{2}. however if i read this [ for all ε < 0, for at least one δ < 0 f([aδ,a+δ]) ⊂ [Lε,L+ε] ] or [ for all ε < 0, for at least one δ < 0 f([aδ,a+δ]) ⊂ f([f^{1}(Lε),f^{1}(L+ε)]) ] assuming f^{1}(Lε) and f^{1}(Lε) is unique but since [ for all ε < 0, for at least one δ < 0 f([aδ,a+δ]) ⊂ f([f^{1}(Lε),f^{1}L+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([aδ,a+δ]) ⊂ [Lε,L+ε] ] which basically says f([aδ,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(domain_{f}) ⊂ R ] assuming f has only one hole (undefined value) and [ for at least one δ < 0 f([aδ,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. 2. i mean , f(B)={f(x)  x is in B } f takes them to members of f(B) ? [ lim x/x at 0 is 1 and lim (x^2x^1)/(x1) 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^2x^1)/(x1) at 1 is 2 and ... ] ] and [ epsilondelta ⇔[ lim x/x at 0 is 1 and lim (x^2x^1)/(x1) at 1 is 2 and ... ] ] and [ [ [ statement_{A} ⇔ statement_{B}] AND [ statement_{A} ⇔ statement_{C} ] ] ⇒ [ statement_{B} ⇔ statement_{C}] ] then [ the deduction ⇔ epsilon delta ] . 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 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 =/= L_{wrong} ] and [ replacing_statement =/> lim f(x) at c = L_{wrong} ] 



#12
Nov3012, 08:23 AM

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PF Gold
P: 9,010

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



#13
Nov3012, 08:53 AM

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P: 9,010

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




#14
Nov3012, 09:05 AM

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#15
Nov3012, 09:41 AM

P: 30

in this statement [ [ 0 < x  a < δ_{1} ⇒ f(x)  L < ε_{1} ] OR [ 0 < x  a < δ_{2} ⇒ f(x)  L < ε_{1} ] OR ... OR [ 0 < x  a < δ_{∞} ⇒ f(x)  L < ε_{1} ] ] or even more better ( hoping that it would priotize your response better ) ignore this whole statement in the previous post [ for all x [ δ_{i} is the ith element of set (0,∞) ] AND [ ε_{i} is the ith element of set (0,∞) ] AND [ [ 0 < x  a < δ_{1} ⇒ f(x)  L < ε_{1} ] OR [ 0 < x  a < δ_{2} ⇒ f(x)  L < ε_{1} ] OR ... OR [ 0 < x  a < δ_{∞} ⇒ f(x)  L < ε_{1} ] ] AND [ [ 0 < x  a < δ_{1} ⇒ f(x)  L < ε_{2} ] OR [ 0 < x  a < δ_{2} ⇒ f(x)  L < ε_{2} ] OR ... OR [ 0 < x  a < δ_{∞} ⇒ f(x)  L < ε_{2} ] ] AND ... [ [ 0 < x  a < δ_{1} ⇒ f(x)  L < ε_{∞} ] OR [ 0 < x  a < δ_{2} ⇒ f(x)  L < ε_{∞} ] OR ... OR [ 0 < x  a < δ_{∞} ⇒ f(x)  L < ε_{∞} ] ] ] => [ lim_{x>a}f(x)=L] 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 never said [ epsilondelta ⇔[ lim x/x at 0 is 1 and lim (x^2x^1)/(x1) at 1 is 2 and ... ] ] is true. i said if you think it is true. 



#16
Nov3012, 10:15 AM

P: 30

to generalize following statement [ lim x/x at 0 is 1 and lim (x^2x^1)/(x1) at 1 is 2 and ... ] [IGNORE] please ignore this in my previous post its not what i wanted to say. [ f(domain_{f}) ⊂ R ] assuming f has only one hole (undefined value) and [ for at least one δ > 0 f([aδ,a+δ]) ⊂ R ] assuming f has any number of hole [/IGNORE] 



#17
Nov3012, 10:59 AM

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#18
Nov3012, 11:36 AM

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P: 779

##(\forall \epsilon \, : \epsilon > 0)(\exists \delta \, : \, \delta > 0)(\forall x)((0 < xa < \delta)\rightarrow(0<f(x)L<\epsilon))## 


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