Infinite tetration and superroot of infinitesimal

[Ivars]

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:
\( \lim_{x\to\infty} x^{(1/x)^{x^{1/x}}}=1 \)

Hi,

I reread the beginning of the thread and it was dx^(1/dx) that was a problem, so lim x^1/x when x-> 0 is undefined in real numbers - that I understood from both jaydfox and Andy.

Since in my understanding infinity is 1/infinitesimal, x-> infinity is perhaps complementary to first one.

When You add another x making x^(1/x)^x You perhaps forcibly drive the expression into the limit You wish as exponent is higher level of infinitesimal than x is infinity. It does not prove (1/dx)^dx=1- these infinitesimals must be of same order.

I chose polar coordinates because there You can see what infinity of self root looks like - it is a unit circle-all 4 expressions tend to it, though from different sides. See attached file. As to self root close to 0, dx^1/dx, 2 of selfroots seem to start at 0, 2-at infinity, depending on combination of signs, so this is still undecided for me.