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## limit : alternate defination

Quote by ato 					 				 				i really dont know you mean by "computer style code" .
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))##

1. using AND,OR comes under logic. it helps to say what we want say to in more flexible form .
I'm pretty sure Fredrik knows what conjunction and disjunction are.

2. using square brackets are needed to show exactly which statements are joined by AND OR =>
We use round brackets actually.

A => B AND C  could either be misunderstood as A => [B AND C] or [ A => B ] AND C
The first one. Conjunction has higher precedence than implication.

Quote by Fredrik

I have no idea what you mean by δ1, ε1, δ2, and so on.
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

lim f(x) at a is L if and only if f([x1,x2]) is real interval for all x1 ,x2   ∈ domainf
update:

lim f(x) at a is L if and only if f([a-δ,a+δ]) is real interval for at least one δ > 0
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 .

Mentor

Quote by ato 					 				 				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.
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.

Quote by ato 					 				 				however i would insist/request you/anyone to give confirmation on
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.

Mentor

Quote by ato 					 				 				nothing => [ [ b =/= L] => [ lim f(c) =/= b ]]

Quote by Mark44 					 				 				I have no idea what you're trying to say with that. It is logically meaningless.

Quote by ato 					 				 				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.
I still have no idea what you mean by this.

This symbol -- => -- is usually taken to mean "implies".
Also, to indicate "not equals" many people write !=, which is notation that comes from the C programming language. Even better is to use ≠, a symbol that is available in the Quick Symbols that appear when you click Go Advanced below the input pane.

You can't write stuff like "replacing_statement => b ≠ L" and expect to be understood, without having said what "replacing_statement" represents.

Quote by ato 					 				 				 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.
What does "lim f(x) at c" mean?

Is this what you mean?
$$\lim_{x \to c} f(x)$$

This limit either exists (and is equal to some number, say L) or it doesn't exist. We don't say "limit of f(x) at c."

This limit can exist whether or not f(c) happens to be defined. If it turns out that f is defined at c, the limit doesn't have to be the same value.

Quote by ato 					 				 				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 ]
This is pretty much gibberish.

Mentor
From your update in post #19

Quote by ato 					 				 				f([0-π,0+π]) is a real interval even if lim f(x) at 0 = 1/2 where f(x) = sin x .
I'm not sure what you mean by f([0-π,0+π]). The sine function maps the interval [##-\pi, \pi##] to the interval [-1, 1].

I don't know what you mean by "lim f(x) at 0 = 1/2". Are there typos in this?

The sine function is continuous for all reals, so
$$\lim_{x \to c} sin(x) = sin(c)$$.

Mentor

Quote by Mark44 					 				 				What does "lim f(x) at c" mean?   Is this what you mean? $$\lim_{x \to c} f(x)$$  This limit either exists (and is equal to some number, say L) or it doesn't exist. We don't say "limit of f(x) at c."
As you know, what we do say is "the limit of f(x) as x goes to c". I've always felt that this is a little bit odd. We're talking about the limit of a function, and that function is denoted by f, not f(x). So there should be a way of saying "the limit of f(x) as x goes to c" without mentioning a dummy variable. I would like to say "the limit of f at c". Obviously, this is to be interpreted as "(the limit of f) at c", not "the limit of (f at c)".

I wouldn't approve of the hybrid "the limit of f(x) at c", because when you mention the dummy variable x, you also have to say that it goes to something.

I may have contributed to some confusion by using the phrase "f has a limit at c if..." (without this explanation) in one of my earlier posts in this thread.

Regarding the =/= stuff that you're quoting in the middle of post #21, it seems to me that what he's saying is just this:
$$\lim_{x\to c}f(x)=L\neq b\ \Rightarrow\ \lim_{x\to c}f(x)\neq b.$$

Hey ato, why not just accept the standard definition of limit? Let's suppose your definition of a limit is indeed correct, in that it allows you to prove all the standard theorems of calculus without introducing any inconsistencies. Even then, why even consider a "new definition" if an older definition already does the job in the most simple possible way, unless you can find some specific application to which your definition is tailored?  I think you might consider accepting the work of:    He was the original creator of the limit definition, and he did it 300 years ago. It's the standard you see in most texts, because his definition is simple yet correct.   BiP

Mentor

Quote by Fredrik 					 				 				As you know, what we do say is "the limit of f(x) as x goes to c". I've always felt that this is a little bit odd. We're talking about the limit of a function, and that function is denoted by f, not f(x). So there should be a way of saying "the limit of f(x) as x goes to c" without mentioning a dummy variable.
I don't see why this should be a consideration. c is a number on the x (typically) axis, and we're working with numbers that are "near" c (and are obviously values on the x-axis).

Quote by Fredrik 					 				 				I would like to say "the limit of f at c". Obviously, this is to be interpreted as "(the limit of f) at c", not "the limit of (f at c)".
I understand what you're saying, but I disagree. If a limit L exists, the closer x is to c, the closer f(x) -- not f -- is to L. What I'm doing here is a sort of paraphrase of the ##\epsilon - \delta## definition.

Quote by Fredrik 					 				 				 I wouldn't approve of the hybrid "the limit of f(x) at c", because when you mention the dummy variable x, you also have to say that it goes to something.   I may have contributed to some confusion by using the phrase "f has a limit at c if..." (without this explanation) in one of my earlier posts in this thread.  Regarding the =/= stuff that you're quoting in the middle of post #21, it seems to me that what he's saying is just this: $$\lim_{x\to c}f(x)=L\neq b\ \Rightarrow\ \lim_{x\to c}f(x)\neq b.$$

Mentor

Quote by Mark44 					 				 				I understand what you're saying, but I disagree. If a limit L exists, the closer x is to c, the closer f(x) -- not f -- is to L. What I'm doing here is a sort of paraphrase of the ##\epsilon - \delta## definition.
Yes, we obviously have to say "f(x)" in a sentence that starts with "the closer x is to c...". I'm not saying that we shouldn't use sentences that start that way. I'm just saying that the phrase "the limit of f at c" makes more sense than "the limit of f(x) as x goes to c", since we're dealing with the limit of a function denoted by f.

Your comment about the epsilon-delta definition made me realize that the reason why people use notation and terminology that mentions a dummy variable is that the definition they're working with mentions a dummy variable. I guess this has some pedagogical advantages.

We could of course use an equivalent definition that doesn't mention a dummy variable: For each open interval A that contains L, there's an open interval B such that c is a member of B and f(B) is a subset of A. Edit: Oops, this is wrong! This definition should say "...and f(B-{c}) is a subset of A".

If this had been the most popular definition, the standard notation would probably be ##\lim_c f## instead of ##\lim_{x\to c}f(x)##.

We could say similar things about the notation ##\int_a^b f(x)\,\mathrm{d}x## for a Riemann integral. It would make at least as much sense to write ##\int_a^b f##, but it's nice to have a notation for the integral that makes it look like a Riemann sum. The notation reminds us of the definition. This is probably also why a notation without dummy variables is totally dominant in the context of Lebesgue integration. In that case, we don't want to be reminded of Riemann sums.

Quote by Mark44 					 				 				From your update in post #19 I'm not sure what you mean by f([0-π,0+π]). The sine function maps the interval [##-\pi, \pi##] to the interval [-1, 1].   I don't know what you mean by "lim f(x) at 0 = 1/2". Are there typos in this?   The sine function is continuous for all reals, so  $$\lim_{x \to c} sin(x) = sin(c)$$.
f([0-π,0+π]) is a set (not function) constructed by taking each and every number from [0-π,0+π] , calculated f(that_number) putting into the set f([0-π,0+π]) .
i borrowed this notation from Fredrick's post , when he used f(B) . if there is anyother more standard and accepted notation, please tell me, i would use that .
whatever the notation is i dont say "f([0-π,0+π]) is a real interval such f-1(-π) and f-1(π) are the endpoints."

what i mean by "lim f(x) at 0 = 1/2" is
limx → 0 f(x) = 1/2 . and what i mean by this is
lets say somehow we came up with a convoluted expression ( called g(x) ) for sin x such as when g(1/2) is calculated we get a undefined form. since limit helps us to define the undefined, we seek limits (my wrong definition here) help . limits says

lim g(x) at 0 is L if and only if f([0-δ,0+δ]) is real interval for at least one δ > 0
lets put L = 1/2 and try to prove required conditions for at least one δ.
as you can see f([0-π,0+π]) is real interval even if limx → 0 g(x) = 1/2 . because the 0 which was needed from g(0) is provided by f(-π) nad f(π) . hence
limx → 0 g(x) = 1/2 is correct . but thats not we want, so i made a change in what i wrote .

lim f(x) at a is L if and only if f([x1,x2]) is real interval for all x1 ,x2 ∈ domainf
so even "f([0-π,0+π]) is real interval " is true,
f([0-x1,0+x2]) is real interval is false for every 0 < x1 < π , 0 < x2 < π and since we could not prove required conditions .
" limx → 0 g(x) = 1/2 " cant be prove true. however i would mention an addendum

limx→af(x) = L ⇔ f([x1,x2]) is real interval for all x1 ,x2 ∈ domainf  and limx→af(x) != L ⇔ f([x1,x2]) is not real interval for at least one x1 and x2 ∈ domainf
so now we can also prove "limx → 0 g(x) != 1/2" true .

Quote by Mark44 					 				 				This is pretty much gibberish.

Quote by fredrick 					 				 				it seems to me that what he's saying is just this: $$\lim_{x\to c}f(x)=L\neq b\ \Rightarrow\ \lim_{x\to c}f(x)\neq b.$$
no thats not i was saying. here you could use "L != b" to prove "limx→c != b" . for example above i did not use the fact "L != b" above in while proving "limx → 0 g(x) != 1/2". now you could replace replacing_statement with my definition and you would see it does not implies what it should not implies . infact it prove that false which is even more better .

Quote by Fredrik 					 				 				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.
what does not contain a or L ? could you elaborate ?

Quote by Bipolarity 					 				 				Hey ato, why not just accept the standard definition of limit? Let's suppose your definition of a limit is indeed correct, in that it allows you to prove all the standard theorems of calculus without introducing any inconsistencies. Even then, why even consider a "new definition" if an older definition already does the job in the most simple possible way, unless you can find some specific application to which your definition is tailored?
i dont understand exactly what epsilon - delta definition is ? i see two version when i read epsilon delta definition .

1.
for all ε > 0, for at least one δ > 0
f([a-δ,a+δ]) ⊂ [L-ε,L+ε]

2.
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

1st version is quite equivalent to my definition . in other words, i dont understand epsilon delta . i understand at least one equivalent version of it.

about its application ? i dont know . i dont expect it to . the definition served its purpose . thats enough for me .but if i find anything (as a result of my_def) in future that does not agree with current calculus i will considered it a failure . i have not yet. if i do i would certainly post here .

Quote by ato 					 				 				f([0-π,0+π]) is a set (not function) constructed by taking each and every number from [0-π,0+π] , calculated f(that_number) putting into the set f([0-π,0+π]) . i borrowed this notation from Fredrick's post , when he used f(B) . if there is anyother more standard and accepted notation, please tell me, i would use that .
Its the correct notation. It just looks strange because you write 0, when most of us would just drop it.

whatever the notation is i dont say "f([0-π,0+π]) is a real interval such f-1(-π) and f-1(π) are the endpoints."
You keep bringing up this concept of "##f([x_1,x_2])## is a real interval". Why? It's not true in general. And its not important when discussing limits.

since limit helps us to define the undefined
Limits do no such thing. If you are talking about indeterminate forms, then that is a different matter.

i dont understand exactly what epsilon - delta definition is ? i see two version when i read epsilon delta definition .  1. for all ε > 0, for at least one δ > 0 f([a-δ,a+δ]) ⊂ [L-ε,L+ε]   <SNIP>  1st version is quite equivalent to my definition .
You seem to have changed your definition multiple times, so I can't keep up. You need to write your definition out in full.

Mentor

Quote by ato 					 				 				what does not contain a or L ? could you elaborate ?
lim f(x) at a is L if and only if f([x1,x2]) is real interval for all x1 ,x2 ∈ domainf
I'm telling you that the first statement
lim f(x) at a is L
can't possibly be equivalent to the second statement
f([x1,x2]) is real interval for all x1 ,x2 ∈ domainf
since the second statement doesn't contain L or a.

Quote by ato 					 				 				i dont understand exactly what epsilon - delta definition is ? i see two version when i read epsilon delta definition .  1. for all ε > 0, for at least one δ > 0 f([a-δ,a+δ]) ⊂ [L-ε,L+ε]   2. 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
If you replace the closed intervals with open intervals (i.e. change f([a-δ,a+δ]) ⊂ [L-ε,L+ε] to f((a-δ,a+δ)) ⊂ (L-ε,L+ε)) in the first definition, you have a definition that's almost the epsilon-delta definition. However, your definition would fail for functions that are defined at a but not continuous there, for example the f defined by f(x)=0 for all x≠a and f(a)=1.

I think I may have contributed to this confusion by posting two definitions where I made this same mistake. I apologize for that.

I think this would work however: "For each ε > 0, there's a δ > 0 such that f((a-δ,a+δ)-{a}) ⊂ (L-ε,L+ε)".

The second definition will at best fail for a lot more functions (even if we replace the closed intervals by open intervals and remove the point a from the first interval). It might also be completely wrong. I haven't thought it through well enough to know. Maybe it works for strictly increasing functions or something.

Also, you can't say something like "assuming f-1(L-ε) and f-1(L-ε) is unique" after the part of the statement that relies on this. You would have to start the definition by saying something like "for all strictly increasing functions f, we define ##\lim_{x\to c}f(x)## as..."

Quote by Fredrik 					 				 				If you replace the closed intervals with open intervals (i.e. change f([a-δ,a+δ]) ⊂ [L-ε,L+ε] to f((a-δ,a+δ)) ⊂ (L-ε,L+ε)), you have a definition that's almost the epsilon-delta definition.However, your definition would fail for functions that are defined at a but not continuous there, for example the f defined by f(x)=0 for all x≠a and f(a)=1.  I think this would work however: "For each ε > 0, there's a δ > 0 such that f((a-δ,a+δ)-{a}) ⊂ (L-ε,L+ε)".
the mistake i found that i mention in post 19, also apply to this too (sorry for not correcting it here). the gist of it is we cannot use ⊂ there. because what i want to say is to change
for all [y1,y2] ⊂ f([a-δ,a+δ]) , y1,y2 ∈ f([a-δ,a+δ]) . alternativily, ⊂ says something about (L-ε,L+ε) not f((a-δ,a+δ)-{a}).
i want function to remain/become continious after choosing a limit for at a point . by continious i mean f(x) increase through all real numbers (it should not skip a number) if x is increased continiously (without skipping number) from x1 to x2 . thats when i say, f([x1,x2] is a real interval . because for a set to be a real interval, it should include all the numbers between any two of its elements .

let me show step by step transition from your definition (above version) to my definition (recent version).

1. "For each ε > 0, there's a δ > 0 such that f((a-δ,a+δ)-{a}) ⊂ (L-ε,L+ε)"
why ⊂ has to go. for example consider f as a monotonically increasing function . so you could a above statement as "For each ε > 0 such that f(((f-1(L-ε)),f-1(L+ε))-{a}) ⊂ (L-ε,L+ε)" but f(((f-1(L-ε)),f-1(L+ε))-{a}) ⊂ (L-ε,L+ε) will always be true irrespective of L .
we need to say is f((a-δ,a+δ)-{a}) is continious set. f((a-δ,a+δ)-{a}) is a real interval .
so we can say
2. "For each δ > 0 such that f((a-δ,a+δ)-{a}) is real interval" or "For at least one δ > 0 such that f((a-δ,a+δ)-{a}) is real interval"
but excluding a is not good idea, because lets say lim f at c is valid . but even then f((a-δ,a+δ)-{a}) would not be a real interval (at least for monotonically increasing functions).
so including a, we can say
3."For each δ > 0 such that f((a-δ,a+δ)) is real interval" or "For at least one δ > 0 such that f((a-δ,a+δ)) is real interval"
lets rule out "For at least one δ > 0 such that f((a-δ,a+δ)) is real interval". as i mention in post 19,
consider f(x) = sin x and we knowingly define lim sinx as x goes to 0 = 1/2, a wrong value . so to prove that this is actually a correct limit we need to prove "f((a-δ,a+δ)) is real interval" for at least one δ. since f([-π,π]) is a real interval so lim sinx as x goes to 0 = 1/2 . but if we include the for each δ clause, we can prevent that so we L_wrong is indeed wrong because f((-π/6,-π/6]) is not real interval .
so lets test
4. "For each δ > 0 such that f((a-δ,a+δ)) is real interval"
you are correct that this version would not give expected limit , lim f(x) as x goes to a for f(x)=0 for all x!=a and f(a)=1 but then traditional limit does not either , in other word traditional limit would say that limit does not exist for f(x) as x goes to a where as this version try to assign a value to f such that f becomes continious. however this version would fail for every x != a. because {0,1} is not a real interval.
lets use this version
5. "For at least one δ > 0, f((a-Δx,a+Δx)) is real interval for all Δx, where 0<Δx<δ"
now thats we can find out limit for all points except a, f(x)=0 for all x!=a and f(a)=1 .

Quote by Fredrik 					 				 				You asked for a comment about the statementlim f(x) at a is L if and only if f([x1,x2]) is real interval for all x1 ,x2 ∈ domainfI'm telling you that the first statementlim f(x) at a is Lcan't possibly be equivalent to the second statementf([x1,x2]) is real interval for all x1 ,x2 ∈ domainfsince the second statement doesn't contain L or a.
why do you think L or a has to be mentaioned ?
do you think that ,
if
statement A : [ anything1 a anthing2 ]
statement B : [ anything3 ]
then [ A <=> B ] is false ?
if yes, why would you think that ? i cant think of any benifit it would give ? or worse it gives wrong results ,like this
[ f is increasing function ] <=> [ for every x1, x2 ∈ domainf [ x1 < x2 <=> f(x1 < x2) ] ] would be false which its not supposed to be ?

or change L in LHS and if RHS does not changes (it does, thats what i say) then let me know i would give more arguements against this .
if RHS does changes then you would notice that correct L, f([x1,x2]) becomes a real interval and vice versa . and thats what i said .

Mentor

Quote by ato 					 				 				why do you think L or a has to be mentaioned ?
OK, there are equivalences where a variable is mentioned in one of the statements but not the other, for example
$$x=1 \text{ and } x\neq 1 \Leftrightarrow 0\neq 0.$$ However, in the case of the two statements we're talking about, it couldn't possibly be more obvious that the two statements are not equivalent. Note for example that given a function f and a number a the truth value (true or false) of the first statement depends on the value of the variable L, but the truth value of the second statement does not.

Ato, you are creating definition in order to solve a problem that was solved a few centuries ago. Your statements may be correct in your head (or maybe not) but do not follow normal practices in logic. Your statement $\lim_{x\rightarrow a}f(x)=L\iff f([x_1,x_2])$ is a real interval for all $x_1,x_2$ on the function's domain is not the definition of a limit. Take the example of $f(x)=\frac{\sin x}{x}$ which has a limit of 1 as $x\rightarrow 0$ but $f([0,\pi])$ is not an interval because $f(0)$ is not defined. Perhaps the idea that you have in your mind is correct, but your notation is wrong.  You should stop trying to come up with a new way to say something before you understand the old way. Learn calculus before you attempt to rewrite it. Even if you are some genius who came up with a better way to define limits, you are not capable of speaking the language of mathematics to get your point across. Go back to your books (or get new ones) and learn how limits work. They do. The definition is good. It really doesn't make sense to discuss this with you until you understand the definition of a limit.  Also, http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/ would make my eyes hurt a little less.

Mentor  I wholeheartedly agree with everything that DrewD said. I just want to add that we have put together a guide for using LaTeX here in the forum. Link.

Blog Entries: 8 				 				 				 				         Recognitions:      Gold Member        Science  Advisor     Staff Emeritus  This thread has gone on for long enough. Ato, I would suggest you to read a good calculus book such as Spivak or Apostol. Limits are very well understood these days. And the definition we have right now works.  Please make some effort to understand our definition. If you do, then we might discuss things again.

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