Since $2^x \not\in O(2^x/x)$, we do not have $O(2^x/x)=O(2^x)$.
But since $x$ rises linearly and $2^x$ exponentially, $2^x/x$ rises almost as fast as $2^x$. Can I somehow express this in Landau notation or the like?
Details: I guess I could just say that both rise exponentially, but since $2^x/x = (2/\sqrt[x]{x})^x$ is asymptotically closer to $2^x$ than $c^x$ is (for a constant $c \neq 2$), I am looking for a notation/solution that is finer grained.
According to this Wikipedia entry, the "soft-O" notation is common in computer science, with $f(n) \in \tilde{O}(g(n))$ being shorthand for $f(n) \in O(g(n) \log^k g(n))$ for some constant $k$. In particular, $\tilde{O}(2^x/x) = \tilde{O}(2^x)$.
von zur Gathen and Gerhard attribute this notation to a 1988 paper of Babai, Luks, and Seress, but I don't know the exact reference. It could well be this one:
SIAM Journal on Computing, 1997, Vol. 26, No. 5 : pp. 1310-1342
Fast Management of Permutation Groups I László Babai, Eugene M. Luks, and Ákos Seress
I found http://ix.cs.uoregon.edu/~luks/fast.pdf which uses $O^\sim\!(g(n))$ rather than $\tilde{O}(g(n))$.