limit related to the Lambert function I am trying to evaluate the following limit 
$$
L=\lim_{x \rightarrow 0^+}\frac{2 \operatorname{W}\left( -{{ e}^{-x-1}}\right)  \left( {{\operatorname{W}\left( -{{e}^{-x-1}}\right) }^{2}}+2 \operatorname{W}\left( -{{ e}^{-x-1}}\right) -2 x+1\right) }{{{\left( \operatorname{W}\left( -{{ e}^{-x-1}}\right) +1\right) }^{3}}}$$
where $W(z)$ is the principal branch of Lambert's function.
The numerical experiments show that it is $\sqrt{2}$ but the l'Hopital's rule does not produce anything useful.
Here is the numerical experiment computed with Maxima: $L(x) - \sqrt{2}$

 A: L'Hopital's rule works.
Note that 
$$\lim_{x\to 0^{+}} W(-\mathrm{e}^{-x-1}) = -1 \tag{1}$$ 
and
$$\frac{\mathrm{d}}{\mathrm{d} x}W(-\mathrm{e}^{-x-1}) = -\frac{W(-\mathrm{e}^{-x-1})}{W(-\mathrm{e}^{-x-1}) + 1}, 
\quad x > 0 \tag{2}$$
where we have used $W'(y) = \frac{W(y)}{y(1+W(y))}$ and the chain rule.
See: https://en.wikipedia.org/wiki/Lambert_W_function
Let
\begin{align}
f(x) &= W(-\mathrm{e}^{-x-1})^2 + 2W(-\mathrm{e}^{-x-1}) - 2x + 1, \\
g(x) &= (W(-\mathrm{e}^{-x-1}) + 1)^3.
\end{align}
We have (noting (2))
\begin{align}
f'(x) &= -2 W(-\mathrm{e}^{-x-1}) - 2, \quad x > 0\\
g'(x) &= -3 (W(-\mathrm{e}^{-x-1}) + 1)W(-\mathrm{e}^{-x-1}), \quad x > 0.
\end{align}
Clearly, $\lim_{x\to 0^{+}} f(x) = 0$ and $\lim_{x\to 0^{+}} g(x) = 0$.
Also, we have (noting (1))
$$\lim_{x\to 0^{+}} \frac{f'}{g'} = \lim_{x\to 0^{+}} \frac{2}{3 W(-\mathrm{e}^{-x-1}) } = -\frac{2}{3}.$$
By L'Hopital's rule, we have $\lim_{x\to 0^{+}} \frac{f}{g} = - \frac{2}{3}$.
Thus, we have (noting (1))
\begin{align}
\lim_{x\to 0^{+}} L &= 2 \cdot \lim_{x\to 0^{+}} W(-\mathrm{e}^{-x-1}) \cdot \lim_{x\to 0^{+}} \frac{f}{g}\\
 &= \frac{4}{3}.
\end{align}
A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
\begin{align} 
L
&=\lim_{x \rightarrow 0^+}
\frac{2 \W(-e^{-x-1})
\left(\W(-e^{-x-1})^2+2 \W(-e^{-x-1}) -2 x+1\right)}
{(\W(-e^{-x-1}) +1)^3}
\tag{1}\label{1}
\\
&=
2
\lim_{x \rightarrow 0^+}\W(-e^{-x-1})
\cdot
\lim_{x \rightarrow 0^+}
\frac{(\W(-e^{-x-1})+1)^2-2 x}
{(\W(-e^{-x-1}) +1)^3}
\tag{2}\label{2}
=-2\cdot L_1
,\\
L_1&=
\lim_{x \rightarrow 0^+}
\frac{(\W(-e^{-x-1})+1)^2-2 x}
{(\W(-e^{-x-1}) +1)^3}
\tag{3}\label{3}
.
\end{align} 
Let 
\begin{align} 
y&=\W(-e^{-x-1})+1
\tag{4}\label{4}
,\\
x&=-(y+\ln(1-y))
\tag{5}\label{5}
,
\end{align} 
$y\rightarrow 0^+$ when $x\rightarrow 0^+$,
so
\begin{align} 
L_1&=
\lim_{y \rightarrow 0^+}
\frac{y^2+2 (y+\ln(1-y))}
{y^3}
\tag{6}\label{6}
\end{align}
Now we can apply the L'Hopital's rule just once:
\begin{align} 
L_1&=
\lim_{y \rightarrow 0^+}
\frac{2y+2-\frac2{1-y}}
{3y^2}
\tag{7}\label{7}
\\
&=
\lim_{y \rightarrow 0^+}
\frac23\cdot\frac1{y-1}
=-\frac23
\tag{8}\label{8}
,
\end{align}
hence
\begin{align} 
L&=-2\cdot L_1
=\frac43
\tag{9}\label{9}
.
\end{align}
$\endgroup$
