# Function grows slower than $\ln(x)$

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?

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

$$f(x)=\sqrt{\ln x}$$

should do, for $x\ge 1$.

• In my opinion this is a much simpler answer than the accepted one. Based on my policy "simpler is better" you have my +1. – Paramanand Singh Aug 8 '15 at 7:52

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

• 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. – Clayton Aug 7 '15 at 21:01
• 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/… – zhw. Aug 7 '15 at 21:16

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.

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.

$\ln(x)$ 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.

• I would say that $\ln \sqrt x$ grows at the same rate as $\ln x$ in terms of asymptotics since their ratio tends to $1/2$. – jdods Aug 7 '15 at 21:18
• $e^{-x}$ doesn't grow. – zhw. Aug 7 '15 at 21:18
• @zwh actually it's negative growth – Zach466920 Aug 7 '15 at 21:19
• @jdods I was looking at the derivative, which does get arbitrarily close to being equal... – Zach466920 Aug 7 '15 at 21:22

$$f(x)=\underbrace{\ln(\ln(\ln(\dots(}_{\lfloor x\rfloor\text{ times}}x)\dots)))$$ 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…)

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 $f(x)$.

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