I am looking for a function $f(x): \Re_{>0} \to \Re_{>0}$ that is monotonically increasing such that
\begin{equation} \tag{1} \label{lim}
\lim_{x \to \infty } \frac{\exp\left\{ f(x) \right\} }{\sqrt{x} f(x)} = C \,, \end{equation}
for some finite, nonzero $C$.
It is easily checked that,
$$ \tag{2} \label{cases} \lim_{x \to \infty} \frac{\exp\left\{ \beta \log(x)^\alpha \right\} }{\sqrt{x} \beta \log(x)^\alpha} = \begin{cases} 0 & \text{if } \alpha < 1 \text{or } \alpha =1 \text{ and } \beta \leq 1/2, \\
\infty & \text{if } \alpha > 1. \end{cases} $$
Hence, my sharpest result so far is $f(x) = \frac{1}{2} \log(x)$.
The question is whether a function that produces a nontrivial limit exists and how it looks like.
Alternatively, I would also be very happy with any rate optimal function $f(x)$ such that for any other function $g(x)$ for which the limit in \eqref{lim} is zero, it also holds that $$ \lim_{x \to \infty} \frac{g(x)}{f(x)} = c \,. $$ for some $c\geq 0$.
Best Answer
You don't need the Lambert function.
$$f(x)=\frac12\log( x) +\log\log (x)+c\\ \frac{\exp(f(x))}{\sqrt{x}f(x)}=\frac{\sqrt{x}\log(x)e^c}{\sqrt{x}f(x)}\\\to2e^c$$