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?
For $x&gt;1$, $x \mapsto \ln(\ln(x))$.
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 $x$

$x^2 + 1 &gt; x \Rightarrow \frac{1}{x^2 + 1}&lt; \frac{1}{x}$

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 $f(x)$.