Limit of Lambert $W$ Product Log is the Natural Log? 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?
$$\lim_{n \rightarrow 0}  n \cdot W\left( \frac{y^{1/n}}{n}\right)= \ln y$$
 A: 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. $\log y$.
A: $\require{begingroup} \begingroup$
$\def\W{\operatorname{W}}$

\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}
$\endgroup$
A: 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   

