Ivars Wrote: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.
dx is used in the context of differentiation and integration. If you use it out of context, nobody knows what you mean. And if you additionally mix it into the theory of hyperreals, that doesnt simplify things. Limits however are well established in mathematics and
\( \lim_{x\downarrow 0} x^{1/x} = \lim_{x\downarrow 0} e^{\ln(x)/x} = 0 \) because \( \ln(x)/x \to -\infty \) for \( x\downarrow 0 \) (\( x\downarrow 0 \) means \( x\to 0 \) while \( x>0 \)).
There is nothing special about it.
These are the other limits:
\( \lim_{x\downarrow 0} x^{1/x^x}= \lim_{x\downarrow 0} e^{\ln(x)/x^x} = 0 \) because \( \lim_{x\downarrow 0 } x^x = 1 \).
\( \lim_{x\downarrow 0} x^{(1/x)^{x^{1/x}}} = 0 \) because \( \lim_{x\downarrow 0} x^{x^{1/x}} = 1 \).
Quote:I chose polar coordinates because there You can see what infinity of self root looks like - it is a unit circleWhat? The limit is a number and not a circle, how can a limit be a geometrical shape?