# 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 am looking for functions that are unbounded and strictly increasing for real $$x>0$$ and grow slower than $$\ln(x)$$

• I think $x\mapsto 1$ "grows" slower than $\ln x$ as $x$ approaches to infinity. Commented Aug 7, 2015 at 21:07
• You probably ought to add that the function is unbounded and strictly increasing. Commented Aug 7, 2015 at 21:35

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. Commented Aug 7, 2015 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.
Commented Aug 7, 2015 at 21:16

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

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. Commented Aug 8, 2015 at 7:52

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.

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.

• I prefer the superlog ( inverse of tetration )
– mick
Commented Nov 11, 2023 at 12:27

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

$\ln(x)$ 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 $1/2$. Commented Aug 7, 2015 at 21:18
• $e^{-x}$ doesn't grow.
– zhw.
Commented Aug 7, 2015 at 21:18
• @zwh actually it's negative growth Commented Aug 7, 2015 at 21:19
• @jdods I was looking at the derivative, which does get arbitrarily close to being equal... Commented Aug 7, 2015 at 21:22

We can get slower functions by substraction, division, functional composition etc.

So by example $$\ln(x) - f(x)$$ or $$\ln(x) / h(x)$$ for suitable $$h(x)$$ and $$f(x)$$.

For instance $$\exp(\sqrt \ln(\ln(x)))$$ will also do.

Or $$\ln(\sqrt x)$$.

Or $$\sqrt\ln(x)$$

Or maybe you like iterative stuff

if $$f(f(x)) = \ln(1+\ln(1+\ln(1+x)))$$

then $$f(x)$$ must grow much slower than $$\ln(x)$$.

Even

$$f(x) = \sum_{n=2}^{\infty}\frac{(\ln(x))^{1/n} - 1}{n^2}$$

grows slower.

Or the more general

$$f(x) = \int_{q}^{\infty}\frac{(\ln(x))^{1/t} - 1}{g(t)} dt$$

for $$q > 1$$ and suitable $$g(t) > 0$$ that satisfies

$$\int_{q}^{\infty} |\frac{1}{g(t)}| dt < \infty.$$

We can use taylors series and differential equations too to get slower functions.

Every branch of math can be used to get slower functions.

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. Commented Aug 7, 2015 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. Commented Aug 7, 2015 at 21:40
• @ShaneT I do not understand the downvotes since it was indeed not explicitly mentioned at the OP. However I recommend just changing your examples to fit the criterium. And maybe the fact you did not change your answer is motivation for the downvotes. I will certainly upvote it if a good edit was made !
– mick
Commented Nov 12, 2023 at 15:14