$\lim_{x \to 0} \frac{x(e^x - 1) + 2(\cos x -1)}{x(1 - \cos x)} = \lim_{x \to 0} \frac{x(e^x - 1)}{x(1 - \cos x)} + \frac{2(\cos x -1)}{x(1 - \cos x)}$

$=\lim_{x \to 0} \frac{(e^x - 1)}{(1 - \cos x)} - \frac{2}{x}$

Applying L'Hospital's rule,

$=\lim_{x \to 0} \frac{(e^x)}{(\sin x)} - \frac{0}{1}$

$=\lim_{x \to 0} \frac{(e^x)}{(\cos x)} = 1$

$=\lim_{x \to 0} \frac{(e^x - 1)}{(1 - \cos x)} - \frac{2}{x}$

Applying L'Hospital's rule,

$=\lim_{x \to 0} \frac{(e^x)}{(\sin x)} - \frac{0}{1}$

$=\lim_{x \to 0} \frac{(e^x)}{(\cos x)} = 1$