The number $$y:=e^{-W_{-1}(-\ln(x))}$$ for $$1< x< e^{e^{-1}}$$ is the solution of the equation $x^y=y$ with $y>e$
How does the function $f(x)=e^{-W_{-1}(-\ln(x))}$ behave for $x\approx 1$ ? In other words, how can I approximate the larger solution of $x^y=y$ for $x>1$ and $x\approx 1$ ?
A very crude approximation seems to be $$\frac{1}{x-1}\ln(\frac{1}{x-1})$$ but surely there is a better asymptotic approximation.
Some notes:
\begin{align} e^{- W_{-1}(-\ln(x))} &= \frac{W_{-1}(- \ln(x))}{- \ln(x)} \\ &\approx - \frac{\ln(\ln(x))}{\ln(x)} + \frac{\ln(- \ln(\ln(x)))}{\ln(x)} - \frac{\ln(- \ln(\ln(x)))}{\ln(x) \, \ln(\ln(x))} \end{align} and \begin{align} \lim_{x \to 1} \, e^{- W_{-1}(-\ln(x))} &= \lim_{x \to 1} \frac{W_{-1}(- \ln(x))}{- \ln(x)} \to \frac{0}{0} \\ &= \lim_{x \to 1} \frac{W'_{-1}(- \ln(x)) \, \left(- \frac{1}{x}\right)}{\left(- \frac{1}{x}\right)} \\ &= \lim_{x \to 1} \, W'_{-1}(- \ln(x)) \\ &= \lim_{x \to 1} \frac{1}{-\ln(x) + e^{W_{-1}(-\ln(x))}} = 1. \end{align}