Let us define $$\log^{(i)}{n}=\log{(\log^{(i-1)}{n})}$$ with this condition: $$i\ge{0}:\log^{(i)}{n}\le1$$ i.e. repeated application of the log function. I wonder what is the result of following limit: $$\lim_{n\to\infty}\lim_{i\to\infty}{\log^{(i)}{n}}=\lim_{n\to\infty}{\log\log\log\cdots\log{n}}$$ I feel like the answer would be a constant number like $C$, is it correct?
Conclusion: the way the problem is composed, it is not possible to have repeated application of $\log$ function.
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I assume you mean $$ f(x) = \ln(x) $$ and ponder about the iteration $$ x_{n+1} = f(x_n) \quad (n \in \mathbb{N}) $$ for some starting value $x_0$. As the graph of $\ln$ does not intersect the graph of the identity function, there is no fixed point $$ x^* = f(x^*) $$ which would exist if $\lim x_n$ existed.
The way you wrote it, the limit does not exist, because for every $n$, there exists some $i$ such that $\log^{(i)}(n) \leq 0$, and thus $\log^{(i+1)}(n)$ does not exist.