[Math] Function grows slower than $\ln(x)$

calculuslimitsreal-analysis

What function grows slower than $\ln(x)$ as $x \rightarrow\infty$? How am I supposed to find it besides just trying finding limits of all known functions?

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For $x>1$, $x \mapsto \ln(\ln(x))$.

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[Math] Function that grows slower than log, using exponentiation instead of multiplication

If you square $\displaystyle 2$, $\displaystyle k$ times, you get $\displaystyle 2^{2^k}$.

Thus you need

$$ 2^{2^k} \ge n$$

i.e.

$$ k \ge \log_2 (\log_2 n)$$

and so you can pick

$$ k = \lceil \log_2 (\log_2 n) \rceil$$

Where $\displaystyle \log_2$ is log to base $\displaystyle 2$.

[Math] ny nonconstant function that grows (at infinity) slower than all iterations of the (natural) logarithm

Yes, just take a function which is equal to $\log x$ on an initial interval such as $[1,K_1]$, then equal to $\log\log x$ on the interval $(K_1,K_2]$, then $\log\log\log x$ etc. where the values $K_1, K_2, \dots$ are chosen sufficiently large to ensure that the function really does tend to infinity. This can always be done, since each of the iterations of the log function does tend to infinity.

This function will not be continuous, but a similar function could certainly be created which is continuous.

This will produce a function which tends to infinity slower than any (fixed) iteration of $\log$. If you want something even slower, then call my function above $\phi (x)$, and repeat my construction using $\phi\phi(x)$ etc on successive intervals.

If this is still not slow enough, then ...

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[Math] Function that grows slower than log, using exponentiation instead of multiplication

[Math] ny nonconstant function that grows (at infinity) slower than all iterations of the (natural) logarithm

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