Answer
$$\frac{1}{4}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{x{e^{ - x}}}}{{2{e^{2x}} - 2}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} x{e^{ - x}} = \left( 0 \right){e^0} = 0 \cr
& \mathop {\lim }\limits_{x \to 0} \left( {2{e^{2x}} - 2} \right) = 2{e^0} - 2 = 2 - 2 = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr
& {\text{denominator}}{\text{:}} \cr
& {\text{for }}x{e^{ - x}} \to {D_x}\left( {x{e^{ - x}}} \right) = x{D_x}\left( {{e^{ - x}}} \right) + {e^{ - x}}{D_x}\left( x \right) = - x{e^{ - x}} + {e^{ - x}} \cr
& {\text{for }}x \to {D_x}\left( {2{e^{2x}} - 2} \right) = 4{e^{2x}} \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{x{e^{ - x}}}}{{2{e^{2x}} - 2}} = \mathop {\lim }\limits_{x \to 0} \frac{{ - x{e^{ - x}} + {e^{ - x}}}}{{4{e^{2x}}}} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{{ - \left( 0 \right){e^{ - 0}} + {e^{ - 0}}}}{{4{e^{2\left( 0 \right)}}}} \cr
& = \frac{1}{4} \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{x{e^{ - x}}}}{{2{e^{2x}} - 2}} = \frac{1}{4} \cr} $$