Prove that $\lim\limits_{\lambda \to\infty}\frac{f(\lambda)}{\lambda/\log_e\lambda}=1$ 
For any real number $\lambda>1$, let $f(\lambda)$ denote the solution to the equation $$x(1+\log_ex)=\lambda.$$
  Prove that $$\lim\limits_{\lambda \to\infty}\frac{f(\lambda)}{\lambda/\log_e\lambda}=1.$$

My Attempt
Let $g(x)=x(1+\log_ex)$. Then $g'(x)=2+\log_ex$ and $g''(x)=\frac{1}{x}$.
Now, $$g\left(\frac{1}{e^2}\right)=-\frac{1}{e^2}$$ is the minimum value of $g(x)$.
But after this, I am not able to get ahead.
 A: Since $$f(\lambda)(1+\ln f(\lambda))=\lambda$$ you have
$$\frac{f(\lambda)\ln\lambda}{\lambda}=\frac{\ln\lambda}{1+\ln f(\lambda)}$$
Moreover $f(\lambda)=\frac{\lambda}{1+\ln f(\lambda)}\leqslant\lambda$ for $\lambda$ large enough. Hence $$\ln(1+\ln f(\lambda))=o(\ln\lambda)$$ and $$\ln f(\lambda)=\ln\lambda-\ln(1+\ln f(\lambda))\sim\ln\lambda$$
Finally $$f(\lambda)\sim\frac{\lambda}{\ln\lambda}$$
A: Let $h(x)=x(1+\log_e(x)):(1,+\infty)\to (1,+\infty)$. Then $h$ is a strictly increasing bijective map with $h^{-1}=f$. Since $\lim_{x\to+\infty} h(x)=+\infty$, by letting $\lambda=h(x)$ we have that
$$\lim_{\lambda \to+\infty}\frac{f(\lambda)}{\frac{\lambda}{\log_e\lambda}}
=\lim_{x \to+\infty}\frac{x}{\frac{h(x)}{\log_e(h(x))}}=
\lim_{x \to+\infty}\frac{\log_e(x)+\log_e(1+\log_e(x))}{1+\log_e(x)}=1.$$
A: This answer uses the Lambert $W$ function which satisfies $z=W(ze^z)$.
Note that \begin{align}f(\lambda)(1+\ln f(\lambda))=\lambda&\implies f(\lambda)=\exp\left({\frac{\lambda}{f(\lambda)}-1}\right)\implies e\lambda=\frac\lambda{f(\lambda)}\exp\left({\frac{\lambda}{f(\lambda)}}\right).\end{align} Taking Lambert $W$ of both sides yields \begin{align}W(e\lambda)=\frac\lambda{f(\lambda)}\implies f(\lambda)=\frac{\lambda}{W(e\lambda)}\end{align} so the limit becomes \begin{align}L=\lim_{\lambda\to\infty}\frac{f(\lambda)}{\lambda/\ln\lambda}=\lim_{\lambda\to\infty}\frac{\lambda/W(e\lambda)}{\lambda/\ln\lambda}=\lim_{\lambda\to\infty}\frac{\ln\lambda}{W(e\lambda)}.\end{align} As both top and bottom diverge to $+\infty$, L'Hopital's rule gives \begin{align}L&=\lim_{\lambda\to\infty}\frac{1/\lambda}{W(e\lambda)/(\lambda W(e\lambda)+\lambda)}=\lim_{\lambda\to\infty}\left(1+\frac1{W(e\lambda)}\right)=1\end{align} as desired.
