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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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    $\begingroup$ I think $x\mapsto 1$ "grows" slower than $\ln x$ as $x$ approaches to infinity. $\endgroup$ – peterwhy Aug 7 '15 at 21:07
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    $\begingroup$ You probably ought to add that the function is unbounded and strictly increasing. $\endgroup$ – marty cohen Aug 7 '15 at 21:35
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$$f(x)=\sqrt{\ln x}$$

should do, for $x\ge 1$.

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  • $\begingroup$ In my opinion this is a much simpler answer than the accepted one. Based on my policy "simpler is better" you have my +1. $\endgroup$ – Paramanand Singh Aug 8 '15 at 7:52
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For $x>1$, $x \mapsto \ln(\ln(x))$.

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    $\begingroup$ To expand on this ever-so-slightly, one can find a function that grows slower than the $n^{\text{th}}$ iteration of the logarithm by just applying another logarithm. $\endgroup$ – Clayton Aug 7 '15 at 21:01
  • $\begingroup$ Given any sequence of functions $g_k\to \infty$, you can always find an $f\to \infty$ such that $f/g_k \to 0$ for every $k.$ See math.stackexchange.com/questions/1370380/… $\endgroup$ – zhw. Aug 7 '15 at 21:16
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If we want to really go to the extreme, we can look at some smooth continuation of the Inverse Ackermann function.

Also, we can look at the inverse of a transexponential function. A function, $f(x)$ is transexponential if there exists some there exists points $p_1, p_2,..., p_n,...$ such that $\forall x > p_1$ $f(x) > e^x$, $\forall x > p_2$, $f(x) < e^{e^x}$, $\forall x > p_3$ $f(x) > e^{e^{e^x}}...$ The inverse of this $f(x)$ will grow very slowly.

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You have to define what you mean by "grows slower than", but I'll assume you want to find a function $f(x)$ such that $\frac{f(x)}{\ln x}\rightarrow 0$ as $x\rightarrow\infty$ where $f(x)\rightarrow\infty$ as well.

L'Hopital's rule is applicable here so that $\displaystyle \lim_{x\rightarrow\infty}\frac{f(x)}{\ln x}=\lim_{x\rightarrow\infty} xf'(x)$. So a function that diverges but whose derivative goes to zero faster than $\frac{1}{x}$ works.

Generally, let $f'(x)=\frac{1}{x}g(x)$ so that $g(x)\rightarrow0$ as $x\rightarrow\infty$ and $\sum \frac{1}{n}g(n)$ diverges, then $f(x)=\int_a^x \frac{1}{t}g(t) dt$ works for a suitably chosen $a$.

For example, $f(x)=\int_2^x \frac{1}{t}\frac{1}{(\ln t)^p} dt$ for $0<p\leq1$ works.

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$\ln(x)$ is monotonic over the relevant interval. So, all one has to do to find a a function that maps $x$ to any function with a derivative less than $1$ at infinity.

In other words,

$$\ln(f(x))$$

Where ${{df} \over {dx}} \lt 1$ as $x$ approaches infinity.

Here are some examples for functions that fit the requirements.

$$f(x)=\sqrt x$$ $$f(x)=e^{-x}$$

You should note that,

$$\ln(f(x)) \gt \ln(x)$$

May hold in the limit. What matters is the rate of growth, not the amount of growth.

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  • $\begingroup$ I would say that $\ln \sqrt x$ grows at the same rate as $\ln x$ in terms of asymptotics since their ratio tends to $1/2$. $\endgroup$ – jdods Aug 7 '15 at 21:18
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    $\begingroup$ $e^{-x}$ doesn't grow. $\endgroup$ – zhw. Aug 7 '15 at 21:18
  • $\begingroup$ @zwh actually it's negative growth $\endgroup$ – Zach466920 Aug 7 '15 at 21:19
  • $\begingroup$ @jdods I was looking at the derivative, which does get arbitrarily close to being equal... $\endgroup$ – Zach466920 Aug 7 '15 at 21:22
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$$f(x)=\underbrace{\ln(\ln(\ln(\dots(}_{\lfloor x\rfloor\text{ times}}x)\dots)))$$ Slightly overkill. (Of course, we could than do $f_2(x)=f(f(\dots(x)\dots))$, and then $f_3(x)=f_2(\dots(x)\dots)$, and then… and then $f_\omega(x)=f_{\lfloor x\rfloor}(x)$, though that last one might not tend to infinity…)

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You could compare the derivatives of two functions, and prove that the derivative of ln(x) is always larger than the derivative of your other function

$f(x) = ln(x)$

$f'(x) = \frac{1}{x}$

$g(x) = arctan(x)$

$g'(x) = \frac{1}{x^2 + 1}$

For all $x$

$x^2 + 1 > x \Rightarrow \frac{1}{x^2 + 1}< \frac{1}{x}$

What this means is that the slope of $f(x)$ is always greater than the slope of $g(x)$ for any given $x$. Which means $g(x)$ is growing slower than $f(x)$.

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    $\begingroup$ But arctan is a bounded function - it never exceeds $\pi/2$. This question is asking about a sequence of unbounded functions. $\endgroup$ – marty cohen Aug 7 '15 at 21:33
  • $\begingroup$ @martycohen Oh, my bad. I didn't realize it specifically asking about an unbounded function since it wasn't written in the OP. $\endgroup$ – Shane T Aug 7 '15 at 21:40

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