Let $\displaystyle f(x) = x^{x^{x^{x^{...}}}} $
What is the maximum value of x such that f(x) is finite?
I wrote a small paper on this for a 2nd year project. The aim was to give a 30 minute easy to follow seminar to our peers.
[Hope it answers your question] Note that this was only the handout for the seminar so it isn't conclusive. Also, it isn't the final draught (cant find it )so there may be a few mistakes
EDIT: Didn't realise which forum this was posted in and so thought you were asking for help
I can see why it works in this case, but simply proving that $\displaystyle f'(x) \rightarrow \infty$ for some $\displaystyle x$ doesn't prove that $\displaystyle f $ is not defined for any values of $\displaystyle x$ greater than this. $\displaystyle \tan(x)$ comes to mind
Sorry, but the function is not defined on this range, so I don't see how it can be continuous. By defining it in the way that you have it is equivalent to the expression involving the Lambert W function that I gave in the PDF, the point of putting it in this form was to show that the function is only defined for certain values of x.
You would need to define the function differently for it to make sense over the range $\displaystyle x \in (e^{1/e},\infty)$
The problem is quite elementary if the approach is correct. By definition for $\displaystyle x>0$ is...
$\displaystyle x^{x}= e^{x \ln x}$
... so that the f(x) you have proposed is...
$\displaystyle f(x)= \lim_{n \rightarrow \infty} e^{x^{n} \ln x}= $
$\displaystyle = 1$ if $\displaystyle 0<x\le 1$
$\displaystyle = +\infty$ if $\displaystyle x>1$
... so that the answer to your question is $\displaystyle x=1$ ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
You seem to have misinterpreted the definition of f.
$\displaystyle \lim_{n \to \infty} e^{x^n \ln{x}} = \lim_{n \to \infty} x^{x^n} \neq x^{x^{x^{x^. . . }}}$
Power towers are read from top to bottom e.g
$\displaystyle 2^{2^{2^2}} = 2^ {2^{4}} = 2^{16} \neq 4^{2^2} = 16^2$
Where on the left hand side of the not equal sign I have calculated correctly, from top to bottom, and on the right hand side I have calculated incorrectly, from bottom to top.
All right!... if power towers are to be read from top to bottom, then we can construct a sequence of functions of $\displaystyle x$ recursively defined as...
$\displaystyle f_{n+1} (x) = x^{f_{n} (x)}$ with $\displaystyle f_{0} (x)=1$
... and we define f(x) as...
$\displaystyle f(x)= \lim_{n \rightarrow \infty} f_{n} (x)$ (1)
Now we have three possibilities...
a) $\displaystyle x>1$
In this case for all $\displaystyle n>0$ is $\displaystyle f_{n} (x) >1$ so that is...
$\displaystyle f(x)= \lim_{n \rightarrow \infty} f_{n} (x) = +\infty$
b) $\displaystyle x=1$
In this case for all $\displaystyle n$ is $\displaystyle f_{n} (1) = 1$ so that is $\displaystyle f(1)=1$
c) $\displaystyle 0 < x < 1$
In this case for all $\displaystyle n>0$ is $\displaystyle 0<f_{n} (x)<1$ but the limit (1) doesn't exist, so that $\displaystyle f(x)$ is undefined. For demonstrate this you can easily demonstrate that ...
$\displaystyle \lim_{x \rightarrow 0+} f_{n} (x) = 1$ for n even, $\displaystyle \lim_{x \rightarrow 0+} f_{n} (x)=0$ for n odd
Anywhere also in this case the answer to the question proposed by R.V. is $\displaystyle x=1$ ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$