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?
Best Answer
According to this post we have that $$W_{-1}(y) = \log(-y) + \log(-\log(-y))+\ldots\qquad \text{for } y\to 0^{-}.$$ Applying this to your problem, we have that \begin{align*}\lim_{x\to +\infty} \frac{f(x)}{x} &= \lim_{x\to +\infty} \frac{2-2x}{x}\cdot \frac{1}{\log x}\,\left[\log\left(\frac{\log x}{(2x-2)\sqrt{x}}\right) + O(\log\log x)\right] \\ &= -2\lim_{x\to +\infty} \frac{1}{\log x}\,\left[\log\log x -\log x^{3/2} + O(\log\log x)\right] \\ &= 3\end{align*} as you claimed.