How can we redefine limits to remove problems with the current definition?

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In summary, the conversation discusses a proposed alternate definition of limits in order to address issues with the current textbook definition. The proposed definition focuses on the values of a function between x and c, rather than just at c. The conversation also addresses misunderstandings and misconceptions about the current definition, such as the use of "as close as possible" and the idea of a limit being equal to 0/0.
  • #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 don't know which when is closest, i don't know when i am wrong . since the only way to prove i am right
is to prove that i cannot be wrong, so i don't 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. let's 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 can't 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.
 
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  • #2
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
ato said:
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 .
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".

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 ?
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].

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

so overall i hate 'possible' and the aim of this post is to remove that 'possible'
Then the purpose of this post is to remove something that isn't there to begin with!

4. the problem 2: since i don't know which when is closest, i don't know when i am wrong . since the only way to prove i am right
is to prove that i cannot be wrong, so i don't know if i am right.
lets say f(c) = 0/0
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 textbook.

but lim (f(c)) = L , then according to current limit defination
nothing => [ [ b =/= L] => [ lim f(c) =/= b ] ]
NO!

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

6. let's define a variable y = ((x^2) - (a^2))/(x-a) AND domain_x = R .
You can't so you are starting with an impossible situation.

but turns out y(a) = 0/0 . however it is not a problem or contradiction or paradox . but you can't 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.
Since your hypothesis is impossible, everything you say here is meaningless.

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

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 ]
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?

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

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

any comment/contradiction is welcome.
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".)
 
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  • #4
ato said:
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.
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.
 
  • #5
-------------

Fredrik said:
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.

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 let's 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 can't prove [ {f(b_1),f(b_2), ... , y_0 , ... , f(b_n)} i interval ]

-> so you can't 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 can't prove otherwise .i think its equivalent to what i said . however it be good if i come across this definition earlier.

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

pwsnafu said:
I implore you to reread (and understand) the actual definition first, because none of your criticisms from points 3 & 4 apply.

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 ?

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

HallsofIvy said:
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".
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.

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

HallsofIvy said:
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 textbook.

"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.

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

HallsofIvy said:
but lim (f(c)) = L , then according to current limit definition
nothing => [ [ b =/= L] => [ lim f(c) =/= b ]]
NO!

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

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

hallsofivy said:
Why not use the standard terminology and call it a continuous variable?
i don't 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 ] .

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

hallsofivy said:
You can't so you are starting with an impossible situation.
how ? but if you have problem with the domain_x , you can just replace it with any other real interval which includes a.

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

halsofivy said:
What does "within x and c" mean? You are the one being vague here.
within x and c <=> from a real interval which include x and c as its endpoint.

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

halsofivy said:
ato said:
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 ]
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?

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.

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

hallsofivy said:
ato said:
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).
I can't make enough grammatical sense out of this statement to comment.
"but should have because"- "should have" what?

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 can't prove [ domain_y_0 is a real interval ]

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

HallsofIvy said:
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].
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.

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

pwsnafu said:
You claim to redefine the concept of "limit of a function"

hallsofivy said:
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!

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 .

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

pwsnafu said:
I recommend you install a spell checker in your browser.

hallsofivy said:
(And, by the way, you are consistently mis-spelling "definition".)

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

-------------
 
  • #6
ato said:
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]
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]

ato said:
so let's 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] ]
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.

ato said:
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 can't prove otherwise .
Textbook definitions do not rely on phrases like "goes to" or "approaches". In fact, those definitions are what assigns a meaning to those phrases.
 
  • #7
ato said:
( and at least 3 out 4 of them has been used multiple times in , stewart calculus 5e)

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.

i am just trying to deduce a equivalent statement that solves the problems , since i think have, i am here for a peer review .

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.
 
  • #8
ato said:
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 ?

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:

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 .

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:
  • #9
micromass said:
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.

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.
 
  • #10
HallsOfIvy said:
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 textbook.

ato said:
"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.

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.


ato said:
but lim (f(c)) = L , then according to current limit definition
nothing => [ [ b =/= L] => [ lim f(c) =/= b ]]
HallsOfIvy said:
NO!
ato said:
i hope you would give something to support this. otherwise it has no meaning to me.

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
nothing => [ [ b =/= L] => [ lim f(c) =/= b ]]
I have no idea what you're trying to say with that. It is logically meaningless.
 
  • #11
when i read for this version of epsilon delta definition
all real ε > 0 there exists a real δ > 0 such that for all x with 0 < |x − a | < δ, we have |ƒ(x) − L| < ε
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
We first pick an ε band around the number L on the y-axis . We then determine a δ band around the number a on the x-axis so that for all
x-values (excluding x=a ) inside the δ band, the corresponding y-values lie inside the ε band.

source : http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html
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 into what we are trying to achieve.

Fredrik said:
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}.

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 }

Fredrik said:
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.

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

pwsnafu said:
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 .
Sorry, but I don't believe you. Your original post made absolutely no attempt to prove that your definition was equivalent to epsilon-delta.

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 ] .


Mark44 said:
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.
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 that's is only what i meant actually as you can see i have not used f(c) = 0/0 anywhere else .

Mark44 said:
ato said:
nothing => [ [ b =/= L] => [ lim f(c) =/= b ]]
I have no idea what you're trying to say with that. It is logically meaningless.

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 ]
 
  • #12
ato said:
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,∞) ]
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.

ato said:
1. b1 is 1st element of B and b2 is 2nd element of B and so on .
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.

ato said:
what do you mean by
f takes them to members of f(B) ?
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.
 
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  • #13
Since you really seem to want to see the definition of "f(x)→L as x→a" in computer code style...

Code:
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 ]
  ]
]
 
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  • #14
ato said:
[ epsilon-delta ⇔[ lim x/x at 0 is 1 and lim (x^2-x^1)/(x-1) at 1 is 2 and ... ] ]

RHS does not imply LHS. All you are doing is looking at the removable singularities of rational functions.
 
  • #15
Fredrik said:
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.
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]

Fredrik said:
Since you really seem to want to see the definition of "f(x)→L as x→a" in computer code style...
i really don't 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

pwsnafu said:
RHS does not imply LHS
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.
 
  • #16
Fredrick said:
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.

i don't want to now because what ever it can say, following statement says too and i just need a confirmation .
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
and i think its better and more easy to translate into 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]
 
  • #17
ato said:
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 ] ]
I have no idea what you mean by δ1, ε1, δ2, and so on.
 
  • #18
ato said:
i really don't 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.
 
  • #19
Fredrik said:
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 don't know , i don't 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:
[strike]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[/strike]
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 .
 
  • #20
ato said:
i mean δ1 means first element of (0,∞) , but i think you are going to ask δ1
or δ2 is. i would say i don't know , i don't 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.

ato said:
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.
 
  • #21
ato said:
nothing => [ [ b =/= L] => [ lim f(c) =/= b ]]
Mark44 said:
I have no idea what you're trying to say with that. It is logically meaningless.
ato said:
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.
ato said:
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.
ato said:
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.
 
  • #22
From your update in post #19
ato said:
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
$$\lim_{x \to c} sin(x) = sin(c)$$.
 
  • #23
Mark44 said:
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:
$$\lim_{x\to c}f(x)=L\neq b\ \Rightarrow\ \lim_{x\to c}f(x)\neq b.$$
 
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  • #24
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:
Augustin-Louis%2BCauchy.jpg


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
Fredrik said:
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).
Fredrik said:
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.
Fredrik said:
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:
$$\lim_{x\to c}f(x)=L\neq b\ \Rightarrow\ \lim_{x\to c}f(x)\neq b.$$
 
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  • #26
Mark44 said:
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.
 
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  • #27
Mark44 said:
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 don't 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 that's 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 " can't 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 .

Mark44 said:
This is pretty much gibberish.
fredrick said:
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 that's 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 .

Fredrik said:
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 ?
Bipolarity said:
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 don't 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 don't understand epsilon delta . i understand at least one equivalent version of it.

about its application ? i don't know . i don't expect it to . the definition served its purpose . that's 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
ato said:
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 don't 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 don't 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.
 
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  • #29
ato said:
what does not contain a or L ? could you elaborate ?
You asked for a comment about the statement
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.

ato said:
i don't 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..."
 
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  • #30
Fredrik said:
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 . that's 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 let's 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 let's 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 that's we can find out limit for all points except a, f(x)=0 for all x!=a and f(a)=1 .

Fredrik said:
You asked for a comment about the statement
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.

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 can't 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, that's 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 that's what i said .
 
  • #31
ato said:
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.
 
  • #32
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
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
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.
 

1. What is the alternate definition of a limit?

The alternate definition of a limit is the mathematical concept that describes the behavior of a function as its input approaches a certain value or point. It is also known as the epsilon-delta definition and is used to formally prove the existence of a limit.

2. How is the alternate definition of a limit different from the traditional definition?

The traditional definition of a limit involves the concept of approaching a value from both sides, while the alternate definition focuses on the distance between the function's output and the desired limit.

3. What is the purpose of using the alternate definition of a limit?

The alternate definition of a limit is used to provide a rigorous and precise definition of the concept, allowing for more accurate mathematical proofs and calculations. It also helps in understanding the behavior of a function near a certain point.

4. Can the alternate definition of a limit be used for all types of functions?

Yes, the alternate definition of a limit can be used for all types of functions, including continuous, discontinuous, and even complex functions. It is a general definition that applies to all types of functions.

5. How is the alternate definition of a limit applied in real-world situations?

The alternate definition of a limit is commonly used in calculus and other branches of mathematics to solve problems involving rates of change, optimization, and approximations. It is also used in physics, engineering, and other scientific fields to model and analyze real-world phenomena.

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