limit related to the Lambert function

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}