[Guest]

I'm not sure if this is a problem or a paradox therefore its up to a moderator to judge.

I was just thinking of this so i want to see what others think. I doubt this has been done before otherwise, great minds think alike.

Prove or disprove that if R4(x)= ln(ln(ln(lnx)))),the limit of Rn(x)as n approaches infinity =0. x could only be one thing in this case so it really wouldn't be a variable.

[superprismatic]

oops i wrote 1 i actually meant 0. lol

No matter what X you start with, iterating the ln function will eventually get to a negative value (right after it falls below 1) and you can't iterate any more because ln is undefined for negative values.

[Guest]

No matter what X you start with, iterating the ln function will eventually get to a negative value (right after it falls below 1) and you can't iterate any more because ln is undefined for negative values.

You could try using ln(x)=ln(-x)+i*pi if x is negative. Not sure what the limit would evaluate to in that case though...

[Guest]

Suppose that there is a number x such that Rn(x) goes to 0 as n goes to infinity. Then for sufficiently large n, the function Rn(x) is arbitrarily small, say less than 1. For such an n, R(n+1)(x)<0 or =0 because the log function is an increasing function. But then R(n+2)(x) is undefined.

[Guest]

exactly. i meant that the only possible x is infinity because any other would be undefined. what i want to know is would x=infinity also be undefined or 0 or infinity?

[Guest]

exactly. i meant that the only possible x is infinity because any other would be undefined. what i want to know is would x=infinity also be undefined or 0 or infinity?

It's been a long time, but I think there are some rules about when you can switch the order of limits that may come into play here.

For any n, Rn(x) -> Inf as x -> Inf. For any x, the limit as n goes to infinity of Rn(x) does not exist as there is always some value N of n such that 0 < RN(x) < 1, and then R(N+1)(x) is undefined.