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