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limit

Type of Math Object: 
Definition
Major Section: 
Reference
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Mathematics Subject Classification

26A06 no label found26B12 no label found54E35 no label found

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is it possible to generalize and include limits to infinity ? and limits that equalize infinity ?

Real Variable Analysis Model of infinite cardinal numbers

Assert that

limit as x --> + infinity of x exists.

Give this limit the name N.

We may, with consistency, identify N with any particular
infinite cardinal number.

I choose to identify N with Aleph-Null.

For any real valued continuous function f,
such that limit as x --> +infinity, is + infinity,

identify f(N) with limit as x --> + infinity of f(x).

Assert that this limit, f(N) exists.

For any two realvalued continued functions f and g,
such that

limit as x --> + infinity of f(x) is + infinity,
and
limit as x --> + infinity of g(x) is + infinity,

Define

f(N) > g(N) if and only if

limit as x --> + infinity of

[ ln(ln(f(x))) / ln(ln(g(x))) ] > 1.

Define f(N) = g(N) if and only if

f(N) is not > g(N) and g(N) is not > g(N).

The operator symbol > is interchangable with the
words "greater than".

Define f(N) < g(N) if and only if g(N) > f(N).

The operator symbol < is interchangable with the
words "less than".

By this definition let's check if the number of prime numbers
is less than N.

The number of prime numbers, by extending the prime number
theorem, is N/ln(N).

limit as x --> + infinity of [ln(ln( N/ln(x) ) )]/[ ln(ln(x))]

= limit as x-->+ infinity of

[ln( ln(x) - ln(ln(x) ) ] / [ ln(ln(x) ]

= by the derivative test, limit as x --> + infinity of

[(1/x - 1/( x ln(x) )/( ln(x) - ln(ln(x)) )]/[ 1/(x ln(x))]

= limit as x --> + infinity of

( ln(x) - 1) / (ln(x) - ln(ln(x)) )

= by the derivative test,

limit as x --> + infinity of

(1/x) / (1/x - 1/(x ln(x) )

= limit as x --> + infinity of

1/(1 - 1/ln(x))

= 1.

Theorem.

If limit as x --> + infinity of f(x)/g(x) = 1,
then f(N) = g(N).

Proof.

If limit as x--> + infinity of f(x)/g(x) = 1
then also,

limit as x --> + infinity of g(x)/f(x) = 1.

Thus f(N) is not > g(N),
and g(N) is not > f(N).

Thus f(N) = g(N)

In this model of comparing infinities,
the number of prime positive integers is
the same as the number of all positive integers.

You may email question and comments directly to me at

kermit@polaris.net

if you wish a response sooner than I would see posts on PlanetMath.

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