From Limits to IMVT

The function $g\,:[1,\infty) \to [1,\infty)$ given by $g(x) = x(1+\ln x)$ is strictly monotonic increasing and bijective, and $f = g^{-1}$ . Putting $x = f(\lambda)$ we see that $\frac{f(\lambda)\ln\lambda}{\lambda} \; = \; \frac{x \ln\lambda}{x(1 + \ln x)} \; = \; \frac{\ln x + \ln(1 + \ln x)}{1 + \ln x} \; = \; \frac{\ln x}{1 + \ln x} + \frac{\ln(1 + \ln x)}{1 + \ln x}$ for $\lambda \ge 1$ . As $\lambda \to \infty$ we see that $x\,,\,1 + \ln x \, \to \infty$ as well, and hence $\lim_{\lambda \to \infty} \frac{f(\lambda)\ln\lambda}{\lambda} \; = \; 1 +0 \; = \; 1$ making the answer $\boxed{\tfrac12}$ .