Linear upper bound for function involving Lambert W I have a function
$$f(x)=\frac{2-2x}{\ln x}W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right)$$
Looking at the plot, it appears somewhat linear in shape and empirically, it seems that we have $\lim_{x\to\infty}f(x)/x=3$. Can this limit be proven, and if so, is there a simpler function that can act as an upper bound for $f(x)$? There is a related problem in which I need to check all integer values less than or equal to $f(x)$ for a given $x$ and I would like to find a simpler function for an upper bound that is still closely fitting enough to not drastically increase computation time.
Update: I have found a function that supplies an upper bound.
$$f(x)\leq2x-2+\frac{(4-4x)[\ln(\ln x)-\ln(x-1)+1-\ln4]}{\ln x}$$
I found this with the observation that $xe^x\geq\frac4{xe^2}$ and thus that $W_{-1}(x)\geq2-\ln4+2\ln(-x)$. Does this help prove the limit and is there a better upper bound still?

Fact 1: It holds that, for all $x > 2$,
$$f(x) \le 3x \cdot \frac{(x - 1)(3\ln x + 2\ln 3 - 2)}{3x\ln x - 2x + 2}.$$
Fact 2: It holds that $f(x) \ge 3x$ for all $x > 2$.
By Facts 1-2, using the squeeze theorem, we have
$$\lim_{x\to \infty} \frac{f(x)}{3x} = 1.$$

Proof of Fact 1:
Let
$$y := \frac{1}{3x} \cdot \frac{2-2x}{\ln x}W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right).$$
Then we have
$$\frac{3xy\ln x}{2 - 2x} = W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right)$$
or
$$\frac{3xy\ln x}{2 - 2x}
\mathrm{exp}\left({\frac{3xy\ln x}{2 - 2x}}\right)
= W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right)\mathrm{exp}\left(W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right)\right)$$
or
$$\frac{3xy\ln x}{2 - 2x}\,
\mathrm{exp}\left({\frac{3xy\ln x}{2 - 2x}}\right) = \frac{\ln x}{(2-2x)\sqrt x}$$
or
$$y\,\mathrm{exp}\left({\frac{3xy\ln x}{2 - 2x}}\right)
= \frac{1}{3x^{3/2}}$$
or
$$\ln y + \frac{3xy\ln x}{2 - 2x} = -\ln (3x^{3/2}). \tag{1}$$
Using $\ln y \le y - 1$ for all $y > 0$, we have
$$y - 1 + \frac{3xy\ln x}{2 - 2x} \ge -\ln (3x^{3/2})$$
which results in
$$y \le \frac{(x - 1)(3\ln x + 2\ln 3 - 2)}{3x\ln x - 2x + 2}.$$
Thus, we have
$$f(x) \le 3x \cdot \frac{(x - 1)(3\ln x + 2\ln 3 - 2)}{3x\ln x - 2x + 2}.$$
We are done.
$\phantom{2}$
Proof of Fact 2:
Let
$$z := W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right).$$
According to definition, we have $z < -1$ and
$$z \mathrm{e}^{z} = \frac{\ln x}{(2-2x)\sqrt x}. $$
It suffices to prove that
$$\frac{1}{3x} \cdot \frac{2-2x}{\ln x} z \ge 1$$
or
$$z \le \frac{3x\ln x}{2 - 2x}.$$
Since $\frac{3x\ln x}{2 - 2x} < -1$ and $z\mapsto z\mathrm{e}^z$ is strictly decreasing on $(-\infty, -1)$, it suffices to prove that
$$z \mathrm{e}^z \ge \frac{3x\ln x}{2 - 2x}\mathrm{exp}\left(\frac{3x\ln x}{2 - 2x}\right)$$
or
$$\frac{\ln x}{(2-2x)\sqrt x} \ge \frac{3x\ln x}{2 - 2x}\mathrm{exp}\left(\frac{3x\ln x}{2 - 2x}\right)$$
or
$$\frac{1}{3x^{3/2}} \le \mathrm{exp}\left(\frac{3x\ln x}{2 - 2x}\right)$$
or
$$-\ln 3 - \frac32\ln x \le \frac{3x\ln x}{2 - 2x}$$
or
$$\ln x\le \frac23(x - 1)\ln 3$$
which is true (easy).
We are done.
