Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{x{e^x}}}{{{e^x} - 1}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} x{e^x} = \left( 0 \right){e^0} = 0 \cr
& \mathop {\lim }\limits_{x \to 0} \left( {{e^x} - 1} \right) = {e^0} - 1 = 1 - 1 = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr
& {\text{denominator}}{\text{.}} \cr
& {\text{for }}{e^{2x}} - 1 \to {D_x}\left( {x{e^x}} \right) = x{D_x}\left( {{e^x}} \right) + {e^x}{D_x}\left( x \right) = x{e^x} + {e^x} \cr
& {\text{for }}x \to {D_x}\left( {{e^x} - 1} \right) = {e^x} \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{x{e^x}}}{{{e^x} - 1}} = \mathop {\lim }\limits_{x \to 0} \frac{{x{e^x} + {e^x}}}{{{e^x}}} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{{\left( 0 \right){e^0} + {e^0}}}{{{e^0}}} \cr
& = \frac{1}{1} \cr
& = 1 \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{x{e^x}}}{{{e^x} - 1}} = 1 \cr} $$
