# limit

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Definition
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## Mathematics Subject Classification

### generalizations

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

### Re: generalizations

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.

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