Infinite tetration and superroot of infinitesimal
Ivars Wrote:r= phi^(1/phi) , r= phi^(1/phi)^phi, r= phi^(1/phi)^phi^(1/phi) etc
...
Now we think that infinity^(1/infinity) which was the beginning of this thread is not very well defined.
...
So why not define the limits using the finite tangent angles? Or their trigonometric functions, e.g tangens? The one of the outgoing/incoming spiral for infinitesimal^(1/infinitesimal), the one of the limit circle at crossing for infinity^1/infinity.

There are options to define those limits , why no one is used?

I dont understand why you invoke a polar graph here, the limits are very well defined and well-known:
\( \lim_{x\to\infty} x^{1/x} = 1 \)
\( \lim_{x\to 0} x^{1/x} = 0 \)
You can already guess it when you look at the (cartesian) graph of the selfroot.

\( \lim_{x\to\infty} x^{(1/x)^x} = \lim_{x\to\infty} x^{1/x^x} = \lim_{x\to\infty} e^{\ln(x)/x^x}=e^{\lim_{x\to\infty}\ln(x)/x^x
}=e^0=1 \)
\( \lim_{x\to\infty} x^{(1/x)^{x^{1/x}}}=\exp\left(\ln(x)\frac{1
}{x^{x^{1/x}}}\right) \)

Here we know that \( x^{1/x}>1 \) for \( x>1 \) and hence \( x^{x^{1/x}}>x \) for \( x>1 \) and hence \( \lim_{x\to\infty}\ln(x)/x^{x^{1/x}}=0 \). Finally again:
\( \lim_{x\to\infty} x^{(1/x)^{x^{1/x}}}=1 \)
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Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by bo198214 - 02/26/2008, 09:26 AM

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