can something exist literally 'infinitely' far away?

Originally posted by humy
I probably should check over this more but I have came up with:

lim { x→∞ } f(x) = L
⇒
∀x ∈ ℝ { f(x)≠L } : ∃a ∈ ℝ : a>x ∧ ¬∃ b ∈ ℝ { b>a } : |f(b) − L| ≥ |f(x) − L|

I am fairly sure you intended ≤ not ≥
[edit]
I take that back. I forgot the ¬ was in effect.
Still it would be neater to have said:
∀ b ∈ ℝ { b>a } : |f(b) − L| < |f(x) − L|
at which point we are getting closer to the Wikipedia definition.

Originally posted by humy
I probably should check over this more but I have came up with:

lim { x→∞ } f(x) = L
⇒
∀x ∈ ℝ { f(x)≠L } : ∃a ∈ ℝ : a>x ∧ ¬∃ b ∈ ℝ { b>a } : |f(b) − L| ≥ |f(x) − L|

Let f(x) = 1/x +1
Let L = 0
Let x=1
Let a=2

∀x ∈ ℝ { f(x)≠L } : ∃a ∈ ℝ : a>x ∧ ¬∃ b ∈ ℝ { b>a } : |f(b) − L| ≥ |f(x) − L|
substituting in the final condition
|f(b) − 0| ≥ |2 − 0|
or |f(b)| ≥ |2|
Since f(x) <2 ∀x ∈ ℝ {x>1} we know that ¬∃ b ∈ ℝ { b>a } |f(b)| ≥ |2|

Therefore the claim lim { x→∞ } 1/x +1 = 0 is satisfied by your definition even though we know the true limit is 1.

Originally posted by twhitehead
Let f(x) = 1/x +1
Let L = 0
Let x=1
Let a=2

∀x ∈ ℝ { f(x)≠L } : ∃a ∈ ℝ : a>x ∧ ¬∃ b ∈ ℝ { b>a } : |f(b) − L| ≥ |f(x) − L|
substituting in the final condition
|f(b) − 0| ≥ |2 − 0|
or |f(b)| ≥ |2|
Since f(x) <2 ∀x ∈ ℝ {x>1} we know that ¬∃ b ∈ ℝ { b>a } |f(b)| ≥ |2|

Therefore the claim lim { x→∞ } 1/x +1 = 0 is satisfied by your definition even though we know the true limit is 1.

So now I try;

lim {x→∞} f(x) = L
means;
" there doesn't exist an x that is such that f(x)≠L and x is such that it isn't true that there both exists an a>x such that f(a) is more than twice as close to L as f(x) is to L but there doesn't exist a b>a that is such that f(b) is further away from L than f(a) is from L ".

That can be expressed as:

lim {x→∞} f(x) = L
⇒
¬∃x ∈ ℝ : f(x)≠L ∧ ¬ ( ∃a ∈ ℝ : a>x ∧ 2*|f(a) − L| < |f(x) − L| ∧ ¬∃ b ∈ ℝ : b>a ∧|f(b) − L| > |f(a) − L| )

Hope I got that right at last.

But I find that just a tricky to take in with there being just two many "¬" there so;

PROVIDING f(x) doesn't equal L throughout some infinite/finite non-zero interval of x;
lim x→∞ f(x) = L
means;
for every x where f(x)≠L , there exists an a>x such that f(a) is more than twice as close to L as f(x) is to L but there doesn't exist a b>a that is such that f(b) is further away from L than f(a) is from L.

That can be expressed as:

lim {x→∞} f(x) = L
⇒
∀x ∈ ℝ { f(x)≠L } : ∃a ∈ ℝ : a>x ∧ 2*|f(a) − L| < |f(x) − L| ∧ ¬∃ b ∈ ℝ : b>a ∧|f(b) − L| > |f(a) − L|
if and only if f(x) doesn't equal L throughout some infinite/finite non-zero interval of x.

Originally posted by humy
Hope I got that right at last.

No. Although all the negatives make it rather hard to analyse.
1. As mentioned before the f(x)≠L is problematic because of horizontal lines such as f(x)=2
2.
Condition 1: x ∈ ℝ : f(x)≠L
Condition 2: a ∈ ℝ : a>x ∧ 2*|f(a) − L| < |f(x) − L|
Condition 3: b ∈ ℝ : b>a ∧|f(b) − L| > |f(a) − L|

Let f(x)=Sin(x)/x
Let L=0
Let x=π-0.1
f(x)=0.1 (approx) ≠L (condition 1 satisfied)

Let a=π
a>x ∧ 2*|0| < |π/2| (condition 2 satisfied)

Let b=5π/2
b>a ∧|1.25 (approx) | > |0.1 approx| (condition 3 satisfied.)

If I read all your negatives correctly, then this violates your definition.

Originally posted by twhitehead
No. Although all the negatives make it rather hard to analyse.
1. As mentioned before the f(x)≠L is problematic because of horizontal lines such as f(x)=2
2.
Condition 1: x ∈ ℝ : f(x)≠L
Condition 2: a ∈ ℝ : a>x ∧ 2*|f(a) − L| < |f(x) − L|
Condition 3: b ∈ ℝ : b>a ∧|f(b) − L| > |f(a) − L|

Let f(x)=Sin(x)/x
Let L=0
Let x=π-0.1
f(x)=0.1 (approx) ...[text shortened]... ion 3 satisfied.)

If I read all your negatives correctly, then this violates your definition.

1. As mentioned before the f(x)≠L is problematic because of horizontal lines such as f(x)=2

Unless I am missing something here, not any more!
Look at the " ¬∃x ∈ ℝ : f(x)≠L ..." bit; for f(x) = 2, there does NOT exist an x such that f(x)≠L thus f(x) = 2 satisfies the condition " ¬∃x ∈ ℝ : f(x)≠L ..." as required (and that renders whatever is to the right of that " ¬∃x ∈ ℝ : f(x)≠L ..." irrelevant) .

But I think I may have made an error of putting the first bracket in the wrong place. Try;


lim { x→∞ } f(x) = L
⇒
¬∃x ∈ ℝ : f(x)≠L ∧ ( ¬∃a ∈ ℝ : a>x ∧ 2*|f(a) − L| < |f(x) − L| ∧ ¬∃ b ∈ ℝ : b>a ∧|f(b) − L| > |f(a) − L| )

I will try and check this later when I have more time although I have a feeling it is still wrong.
I am close to giving up this exercise.

Originally posted by humy
Unless I am missing something here, not any more!
Look at the " ¬∃x ∈ ℝ : f(x)≠L ..." bit; for f(x) = 2, there does NOT exist an x such that f(x)≠L thus f(x) = 2 satisfies the condition " ¬∃x ∈ ℝ : f(x)≠L ..." as required (and that renders whatever is to the right of that " ¬∃x ∈ ℝ : f(x)≠L ..." irrelevant) .

That is why I say 'problematic' not 'wrong'. It just worries me that for some functions your conditions become a little too trivial.

I also find the use of 'twice' to be somewhat arbitrary.

Originally posted by humy
But I think I may have made an error of putting the first bracket in the wrong place.

I have shown that A ∧ B ∧ C
I am fairly sure this contradicts
¬A ∧ (¬B ∧ ¬C)
but all the ¬'s are confusing me.

OK, I give up on independently defining it for myself and will now just take it from the link:


https://en.wikipedia.org/wiki/Limit_of_a_function#Functions_of_a_single_variable
"...
Functions of a single variable

Suppose f : R → R is defined on the real line and p,L ∈ R. It is said the limit of f, as x approaches p, is L and written

lim {x → p} f ( x ) = L

if the following property holds:

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.

The value of the limit does not depend on the value of f(p), nor even that p be in the domain of f.
..."

So, using formal notation, I can write that as:

lim {x→p} f(x) = L
⇒
∀ε ∈ ℝ{>0} ∃δ ∈ ℝ{>0} : ∀x ∈ ℝ{>0} 0 < | x − p | < δ ⇒ | f(x) − L | < ε

(ℝ{>0} is the standard notation for the set of positive real non-zero numbers )

?

I am having difficulty getting the full meaning of that expression assuming I have written it down correct.
Does it really work as required?

Originally posted by humy
So, using formal notation, I can write that as:

lim {x→p} f(x) = L
⇒
∀ε ∈ ℝ{>0} ∃δ ∈ ℝ{>0} : ∀x ∈ ℝ{>0} 0 < | x − p | < δ ⇒ | f(x) − L | < ε

(ℝ{>0} is the standard notation for the set of positive real non-zero numbers )

?

I am having difficulty getting the full meaning of that expression assuming I have written it down correct.
Does it really work as required?

One correction: x doesn't have to be >0.

The meaning is:
For any interval ε around the limit, there is always an interval δ around p for which the function stays within the interval ε around the limit. Note that the function doesn't have to have the value of the limit at p, so the interval δ can be thought of as two open intervals on either side of p. ε on the other hand isn't split in this way.
https://en.wikipedia.org/wiki/Limit_of_a_function#Deleted_versus_non-deleted_limits

Originally posted by twhitehead
One correction: x doesn't have to be >0.

So it should be:

lim x→p f(x) = L
⇒
∀ε ∈ ℝ>0 ∃δ ∈ ℝ>0 : ∀x ∈ ℝ 0 < | x − p | < δ ⇒ | f(x) − L | < ε

I have come up with a formula for defining the limit.
Disadvantage; it only works for certain types of none-oscillating functions.
Advantage; very easily to understand its meaning.
For only those functions of x that are such that they are always increasing with x i.e. if a<b then we necessarily have f(a) < f(b), and that condition can be written as; " if for function f ; ∀a,b ∈ ℝ{a<b}, f(a) < f(b) " we simply have:


if for function f ; ∀a,b ∈ ℝ{a<b}, f(a) < f(b)
then
(
lim {x→∞} f(x) = L
⇒
∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ 2*|f(y) − L| < |f(x) − L|
)

The "2" is arbitrary but don't see how to avoid that.

Formula correct?

Originally posted by humy
Formula correct?

I believe it works for the limited class of functions you have chosen. I don't see any real benefits though.

For bonus points, instead of 2, use n. What is the smallest value of n for which it works?

Originally posted by twhitehead
I


... What is the smallest value of n for which it works?

2 (assuming n is a natural number else, for reals, there is no smallest. We cannot have n=1 else you can increase L for the same function and it would still be defined as the limit! Pity )

Originally posted by humy
2 (assuming n is a natural number else, for reals, there is no smallest. We cannot have n=1 else you can increase L for the same function and it would still be defined as the limit! Pity )

1 is a real, so your claim that for reals there is no smallest doesn't hold.

Originally posted by twhitehead
1 is a real, so your claim that for reals there is no smallest doesn't hold.

But n=1 in:

if for function f ; ∀a,b ∈ ℝ{a<b}, f(a) < f(b)
then
(
lim {x→∞} f(x) = L
⇒
∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ 1*|f(y) − L| < |f(x) − L|
)

doesn't work so there is no smallest real that works for that.