can something exist literally 'infinitely' far away?

Originally posted by humy
doesn't work so there is no smallest real that works for that.

I am not sure what you mean. What about 1.1?

Originally posted by twhitehead
I am not sure what you mean. What about 1.1?

What about it?
1.1 is not the smallest number larger than 1, which is what the 'smallest' would mean in this very narrow context of the constant n that is required in my formula for it to work.
That is because there is no smallest number larger than 1.

Remember, for my formula to work, n in;

∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ n*|f(y) − L| < |f(x) − L|

must be some real number (any real number) larger than 1.
Whatever number you give larger than 1, I can give a number smaller than it but still larger than 1 thus there is no smallest number larger than 1; at least none is smallest if you are talking about all the reals and not restricting it to just the natural numbers.

Originally posted by humy
What about it?
1.1 is not the smallest number larger than 1, which is what the 'smallest' would mean in this very narrow context of the constant n that is required in my formula for it to work.
That is because there is no smallest number larger than 1.

OK, I wasn't sure what you meant but now I see.
So you could actually put in n in your definition and say ∀n ∈ ℝ {n>1}
I am not sure if that improves the arbitrary 2, but then the whole definition has such a narrow scope that its not that important to make it perfect.

Just for fun, because you know that f(x)<L and f(y)<L you can drop the absolutes and rearrange to get:
lim {x→∞} f(x) = L
⇒
∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ f(y)> (L+f(x))/2

Originally posted by twhitehead
Just for fun, because you know that f(x)<L and f(y)<L you can drop the absolutes and rearrange to get:
lim {x→∞} f(x) = L
⇒
∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ f(y)> (L+f(x))/2

well spotted!

Originally posted by twhitehead
Just for fun, because you know that f(x)<L and f(y)<L you can drop the absolutes and rearrange to get:
lim {x→∞} f(x) = L
⇒
∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ f(y)> (L+f(x))/2

You have just giving me an idea how to modify that to make it work for a greater range of functions including some oscillating ones:

if for function f ; lim {x→∞} f(x) = L ∧ ∀y ∈ ℝ : f(y) < L
then
∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ ∀z ∈ ℝ{z>y} : f(z) > (L+f(x))/2

which I presume could be written more simply as:

lim {x→∞} f(x) = L ∧ ∀y ∈ ℝ : f(y) < L
⇒
∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ ∀z ∈ ℝz>y : f(z) > (L+f(x))/2

" (L+f(x))/2 " can be interpreted as simply meaning "the average of L and f(x)".

Originally posted by humy
lim {x→∞} f(x) = L ∧ ∀y ∈ ℝ : f(y) < L
⇒
∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ ∀z ∈ ℝz>y : f(z) > (L+f(x))/2

" (L+f(x))/2 " can be interpreted as simply meaning "the average of L and f(x)".

I don't understand what y is doing, especially in the second line where it is >x and <z but otherwise appears to serve no purpose.
In the first line I believe you could use x again without issue.

Originally posted by twhitehead
I don't understand what y is doing, especially in the second line where it is >x and <z but otherwise appears to serve no purpose.
In the first line I believe you could use x again without issue.

Oh yes. So, improvement:

lim {x→∞} f(x) = L ∧ ∀x ∈ ℝ : f(x) < L
⇒
∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ ∀y : f(y) > (L+f(x))/2

-but I have got rid of z instead of y in the second line.

I thought of " ∀y : " last time but somehow that looked a bit odd to me without a " ∈ ..." after it as I don't ever recall seeing it used that way which is why I added the extra variable z last time. But, now I think about it, I think just writing " ∀y : " there makes sense as I have already shown "y ∈ ℝ : " before it ?
Anyone? Writing " ∀y : ... " is permitted?

Originally posted by humy
Anyone? Writing " ∀y : ... " is permitted?

Actually, my bad. I think the x,y and z were necessary in the second line. I don't think the y in the first was necessary. Certainly, as it stands it doesn't make sense.

Originally posted by twhitehead
Actually, my bad. I think the x,y and z were necessary in the second line. I don't think the y in the first was necessary. Certainly, as it stands it doesn't make sense.

So it should be:

lim {x→∞} f(x) = L ∧ ∀x ∈ ℝ : f(x) < L
⇒
∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ ∀z ∈ ℝ{z>y} : f(z) > (L+f(x))/2

also just noticed I also made the error of forgetting to indicate with curly brackets that "z>y" is a subscript last time. Having it written as " ℝz>y " there is bad notation.

Originally posted by humy
So it should be:

lim {x→∞} f(x) = L ∧ ∀x ∈ ℝ : f(x) < L
⇒
∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ ∀z ∈ ℝ{z>y} : f(z) > (L+f(x))/2

also just noticed I also made the error of forgetting to indicate with curly brackets that "z>y" is a subscript last time. Having it written as " ℝz>y " there is bad notation.

Miner point, but; Just noticed that it would be more consistent with conventional notation to write that subscript as "z>y" as just ">y" thus the formula should be;

lim {x→∞} f(x) = L ∧ ∀x ∈ ℝ : f(x) < L
⇒
∀x ∈ ℝ : ∃y ∈ ℝ : y>x ∧ ∀z ∈ ℝ{>y} : f(z) > (L+f(x))/2

There, that should be perfect now.

This below is supposed to be a definition of limit for all functions of x that don't equal a constant for all x. That would thus exclude functions such as f(x) = 2 etc;

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

(the f(x) ∈ ℝ and L ∈ ℝ parts excludes L = ∞ and f(x) = ∞ )
Does that work for all possible allowed functions?

If we made x = ∞, would the expression of, say;

x < 2*x

be 'false'?
Or would that be neither 'true' or 'false' but rather 'meaningless' as in 'nonsense'?

Originally posted by humy
If we made x = ∞, would the expression of, say;

x < 2*x

be 'false'?
Or would that be neither 'true' or 'false' but rather 'meaningless' as in 'nonsense'?

It depends on what you mean by the symbol ∞.
If you mean 'the Real number infinity', then you are making an error in the first expression as infinity isn't a number. For other meanings / definitions of the symbol, it will depend on the definition as to whether the initial expression makes sense and whether or not the later expressions may be evaluated.

Originally posted by twhitehead
It depends on what you mean by the symbol ∞.
If you mean 'the Real number infinity', then you are making an error in the first expression as infinity isn't a number. For other meanings / definitions of the symbol, it will depend on the definition as to whether the initial expression makes sense and whether or not the later expressions may be evaluated.

perhaps a better question is does either

2*∞ = ∞ is true

or

2*∞ > ∞ is false

make total sense in formal logic?