Does there exist any unbounded above function $f(x)$ such that $f(x)<\log(x)$ for all $x>M$ Does there exist any unbounded above function $f: \mathbb{R} \to \mathbb{R}$ such that there is some $M > 0$ such that $f(x)<\log(x)$ for all $x>M$?
Mainly I observed the fact that $\log(x)$ has slope $0$ at infinity. 
Let 
$g(x)=\log(x)$
$g'(x)= {1\over x}$
$\lim\limits_{x\to \infty}g'(x)=0$
So I think $\log(x)$ is the ultimate boundary. So every unbounded function must have a slope $\ge$ slope of $\log(x) $ at infinity 
 A: $f(x) = \log (x) - 1$ should do nicely.
Other possibilities:


*

*$\log(\log(x))$, or $\log(\log(\log(x)))$ and so on

*$\log(x) - C$ for some constant $C>0$

*$\log(\sqrt{x})$

*$\log(f(x))$ where $f(x)$ is any function bounded above and growing slower than $x\mapsto x$



If you want a function with a non-zero slope, you can take
$$f(x) = (\sin(x) + 2)\cdot \frac{\log(x)}{3} - 1$$
This function is obviously always smaller than $\log x$ and is also unbounded at $\infty$.
Its derivative is
$$f'(x) = \cos x\frac{\log x}{3} +\frac{1}{3x}(\sin x + 2)$$
so the slope is not zero and does not approach zero at $\infty$.
A: Or if you want something growing "a lot" slower, take $f(x) = \log \log x$. (For $x$ large enough.)
A: Just like there is no minimal positive real number, there is no minimal rate of unbounded growth. If $\def\R{\Bbb R}f:(a,+\infty)\to\R$ is any increasing unbounded function, then $x\to f(\log x)$ is a function also an increasing and unbounded function $(\exp(a),+\infty)\to\R$ that is dominated by it.
This construction does not quite guarantee an asymptotically slower rate of growth, but you'll have to work hard to construct to construct a function for which it doesn't (something like inverse tetration). But even then there is asymptotically (much!) slower still, like the inverse Ackermann function.
