$f(x) = O(x^{-1})$ as $x \to \infty$ implies that $f(x) = o(x^{-1+\delta})$ for any $\delta > 0$?

Question

Suppose $f(x) = O(x^{-1})$ as $x \to \infty$. What can we say about $f(x)$ in terms of small o? I know we can say $f(x) = o(1)$ or $f(x) = o(x^{-1/2})$ for example. It seems in fact that we could say $f(x) = o(x^{-1+\delta})$ for any $\delta > 0$. Do you agree?