Does $\ln(\frac{2x}{\ln(2x+1)})\sim W(x)$?

Question

The function $f(x)= \ln(\frac{2x}{ln(2x+1)})$ when plotted is similar to the plot of the Lambert W function. The Wolfram Alpha says that the limit $\lim_{x\rightarrow\infty} \frac{W(x)}{f(x)}=1$, but on the graph it seems that the limit doesn't reach one. Someone can explain if the limit is really equal to $1$ or WFA is incorrect?

An approach is to consider the function $y=\frac{W(x)}{f(x)}$ and see that $$yf(x)e^{yf(x)}=x$$

We can do the limit as $x\rightarrow\infty$ but it seems to make the problem a little bit too complex.

I will be grateful if someone can help me.

Answer 1

For $x \to \infty$ is $$ \begin{align} f(x) &= \ln(2) + \ln(x) - \ln(\ln(2x+1)) \\ &= \ln(2) + \ln(x) - \ln\left(\ln(2) + \ln(x) + \ln\left(1+\frac{1}{2x}\right)\right) \\ &= \ln(2) + \ln(x) - \ln\bigl(\ln(x) + O(1)\bigr) \\ &= \ln(x) - \ln(\ln(x)) + \ln(2) + o(1) \end{align} $$ and (see here) $$ W(x) = \ln(x) - \ln(\ln(x)) +o(1) $$ so that indeed $f(x)/W(x) \to 1$ for $x \to \infty$.

Answer 2

Asymptotically $$\log \left(\frac{2 x}{\log (2 x+1)}\right)-W(x)=\log \left(\frac{2 \log (x)}{\log (2 x)}\right)$$

Considering the norm $$\Phi(a)=\int_0^{10^{10}} \Bigg(\log \left(\frac{2 x}{\log (2 x+1)}\right)-W(ax)\Bigg)^2\,dx$$ it is minimum for $a\sim \frac 74$. Over this range, the match is quite good.

Simpler is $$\log \left(\frac{2 x}{\log (2 x+1)}\right) \sim W(x) +\frac 12$$

Asymptotically $$\log \left(\frac{2 x}{\log (2 x+1)}\right)-\left(W(x)+\frac 12\right)=\log \left(\frac{2 \log (x)}{\sqrt{e} \log (2 x)}\right)$$

Similarly, replacing the $2$ by $\frac 76$ significantly improves.

In any manner, this leads to good approximations of $W(x)$ for large values of the argument and this could be useful.