Function that gives nontrivial limit of $\frac{\exp\left\{ f(x) \right\} }{\sqrt{x} f(x)}$

Question

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 $c\geq 0$.

Answer 1

You don't need the Lambert function.

$$f(x)=\frac12\log( x) +\log\log (x)+c\\ \frac{\exp(f(x))}{\sqrt{x}f(x)}=\frac{\sqrt{x}\log(x)e^c}{\sqrt{x}f(x)}\\\to2e^c$$

Answer 2

For any fixed positive $C$, you can take $$ f(x) = - W_{ - 1}\! \left( { - \frac{1}{{\sqrt {C^2x + {\rm e}^2 } }}} \right) $$ with $W_{-1}$ being one of the branches of the Lambert $W$-function. This $f(x)$ increases monotonically for $x>0$ and is positive valued. Then $$ \frac{{\exp (f(x))}}{{\sqrt x f(x)}} = \frac{{\sqrt {C^2 x + {\rm e}^2 } }}{{\sqrt x }} \to C $$ as $x\to +\infty$.

Addendum. For any fixed $C>0$, the function $$ f(x) = \log ( \sqrt {C^2x + {\rm e}^2 } ) + \log \log (\sqrt {C^2 x + {\rm e}^2} ) $$ is positive valued and monotonically increasing for $x>0$, and satisfies the limit condition.