You know that $e^x$ has no vertical asymptote, but the horizontal asymptote 0; and that $\log(x)$ has no horizontal asymptote but vertical asymptote 0.
I was wondering, if there is a monotone unbounded continuously differentiable function in closed-form, which has no asymptote (not even an oblique one, so also no rotated hyperbola).
In particular, I was searching for a function with $$f(x)< x\,\, \forall\, x \setminus \{0\},\\ f'(x)>0,\\ f''(x)<0$$ for all $x$, as well as $f(0)=0, f'(0)=1$.
The closest I got was this $$ f(x)=\begin{cases}\log(x+1) &,x\ge0\\ 1-e^{-x} &,x<0\end{cases},$$ but this is not closed-form and has a discontinuous 3rd derivative. At first I thought that wouldn't be that difficult, but it's apparently harder than I thought... Does anyone has an idea?
For a monotone unbounded real-analytic function with no asymptotes, you might try $x^3$.
But if you want one that is concave, you might try
$$ f(x) = -{\frac {5\,{x}^{2}{{\rm e}^{-x}}}{16+16\,{{\rm e}^{-x}}}}+{\frac {x}{ \sqrt [4]{{x}^{2}+1}}}$$