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Limit : alternate defination 
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#19
Nov3012, 11:51 AM

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or δ_{2} 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 δ_{1} and δ_{2} exist and using them in a statement is not a problem. however i would insist/request you/anyone to give confirmation on f([0π,0+π]) is a real interval even if lim f(x) at 0 = 1/2 where f(x) = sin x . 


#20
Nov3012, 12:16 PM

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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 epsilondelta definition is saying in a way that you're more comfortable with. And the epsilondelta definition doesn't mention any sequences. 


#21
Nov3012, 01:33 PM

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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. 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. 


#22
Nov3012, 03:42 PM

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From your update in post #19
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 $$\lim_{x \to c} sin(x) = sin(c)$$. 


#23
Nov3012, 04:41 PM

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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: $$\lim_{x\to c}f(x)=L\neq b\ \Rightarrow\ \lim_{x\to c}f(x)\neq b.$$ 


#24
Nov3012, 05:12 PM

P: 783

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 


#25
Nov3012, 05:39 PM

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#26
Dec112, 12:39 AM

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Your comment about the epsilondelta 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. 


#27
Dec112, 02:17 AM

P: 30

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([0π,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 lim_{x → 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 as you can see f([0π,0+π]) is real interval even if lim_{x → 0} g(x) = 1/2 . because the 0 which was needed from g(0) is provided by f(π) nad f(π) . hence lim_{x → 0} g(x) = 1/2 is correct . but thats not we want, so i made a change in what i wrote . f([0x_{1},0+x_{2}]) is real interval is false for every 0 < x_{1} < π , 0 < x_{2} < π and since we could not prove required conditions . " lim_{x → 0} g(x) = 1/2 " cant be prove true. however i would mention an addendum 1. for all ε > 0, for at least one δ > 0 f([aδ,a+δ]) ⊂ [Lε,L+ε] 2. 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 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 . 


#28
Dec112, 02:55 AM

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#29
Dec112, 03:17 AM

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lim f(x) at a is L if and only if f([x_{1},x_{2}]) is real interval for all x_{1} ,x_{2} ∈ domain_{f}I'm telling you that the first statement lim f(x) at a is Lcan't possibly be equivalent to the second statement f([x_{1},x_{2}]) is real interval for all x_{1} ,x_{2} ∈ domain_{f}since the second statement doesn't contain L or a. 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((aδ,a+δ){a}) ⊂ (Lε,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}(Lε) and f^{1}(Lε) 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..." 


#30
Dec212, 09:00 AM

P: 30

for all [y_{1},y_{2}] ⊂ f([aδ,a+δ]) , y_{1},y_{2} ∈ f([aδ,a+δ]) . alternativily, ⊂ says something about (Lε,L+ε) not f((aδ,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 x_{1} to x_{2} . thats when i say, f([x_{1},x_{2}] 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((aδ,a+δ){a}) ⊂ (Lε,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}(Lε)),f^{1}(L+ε)){a}) ⊂ (Lε,L+ε)" but f(((f^{1}(Lε)),f^{1}(L+ε)){a}) ⊂ (Lε,L+ε) will always be true irrespective of L . we need to say is f((aδ,a+δ){a}) is continious set. f((aδ,a+δ){a}) is a real interval . so we can say 2. "For each δ > 0 such that f((aδ,a+δ){a}) is real interval" or "For at least one δ > 0 such that f((aδ,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((aδ,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((aδ,a+δ)) is real interval" or "For at least one δ > 0 such that f((aδ,a+δ)) is real interval" lets rule out "For at least one δ > 0 such that f((aδ,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((aδ,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((aδ,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((aΔx,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 . do you think that , if statement A : [ anything_{1} a anthing_{2} ] statement B : [ anything_{3} ] 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 x_{1}, x_{2} ∈ domain_{f} [ x_{1} < x_{2} <=> f(x_{1} < x_{2}) ] ] 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([x_{1},x_{2}]) becomes a real interval and vice versa . and thats what i said . 


#31
Dec212, 11:06 AM

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$$x=1 \text{ and } x\neq 1 \Leftrightarrow 0\neq 0.$$ 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. 


#32
Dec212, 11:25 AM

P: 460

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 [itex]\lim_{x\rightarrow a}f(x)=L\iff f([x_1,x_2])[/itex] is a real interval for all [itex]x_1,x_2[/itex] on the function's domain is not the definition of a limit. Take the example of [itex]f(x)=\frac{\sin x}{x}[/itex] which has a limit of 1 as [itex]x\rightarrow 0[/itex] but [itex]f([0,\pi])[/itex] is not an interval because [itex]f(0)[/itex] 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. 


#33
Dec212, 12:10 PM

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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.



#34
Dec212, 12:26 PM

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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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