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