Limit of Lambert $W$ Product Log is the Natural Log?

Question

In solving this equation $\large y=x^ne^x$ I get the result that $$n \cdot W\left( \frac{y^{1/n}}{n}\right)=x $$

So now it is apparent to me that when $n=0$ you would simply get $\ln(y)=x$ by normal methods. But is there any way to show that the limit this is true as well?

$$\lim_{n \rightarrow 0} n \cdot W\left( \frac{y^{1/n}}{n}\right)= \ln y$$

Answer 1

On any compact interval of $\mathbb{R}^+$ the sequence of continuous and monotonic functions given by $$ f_m(x) = x^{1/m} e^{x} $$ converges uniformly towards $e^x$, hence, for a given $y\in\mathbb{R}^+$, the sequence $\{x_m\}_{m\in\mathbb{N}^+}$ of the solutions of $f_m(x)=y$ converges towards the solution of $e^x = y$, i.e. $\log y$.

Answer 2

$\require{begingroup} \begingroup$ $\def\W{\operatorname{W}}$

\begin{align} L&=\lim_{n \rightarrow 0} n \cdot \W\left(\tfrac1n y^{1/n}\right).\tag{1}\label{1}\end{align}

With the L'Hospital's rule in mind we have:

\begin{align} L &=\lim_{n \to 0} \frac{\W\left( \tfrac1n\exp(\tfrac1n\ln y)\right)}{1/n} =\lim_{n \to 0} \frac{f(n)}{g(n)} \tag{2}\label{2} ,\\ f(n)&=\W\left( \tfrac1n\exp(\tfrac1n\ln y)\right) ,\quad g(n)=1/n ,\\ \lim_{n \to 0}f(n) &= \lim_{n \to 0}g(n) =\infty ,\\ f'(n)&= -\frac1{n^2}\cdot \frac{\W\left( \tfrac1n\exp(\tfrac1n\ln y)\right)}{1+\W\left( \tfrac1n\exp(\tfrac1n\ln y)\right)} \cdot(n+\ln y) ,\\ g'(n)&= -\frac1{n^2} ,\\ L&= \lim_{n \to 0} \frac{f'(n)}{g'(n)} = \left(1+\W\left( \tfrac1n\exp(\tfrac1n\ln y)\right)^{-1} \right)^{-1} \cdot(n+\ln y) =\ln y . \end{align}

$\endgroup$

Answer 3

Up t0 Maple 2021, this is not true in view of

limit(n*LambertW(2^(1/n)/n), n = 0, right);
                         ln(2)

limit(n*LambertW(2^(1/n)/n), n = 0, left);
                           0
eval(n*LambertW(2^(1/n)/n), n = -0.01);
                                 -31
                   7.888609052 10