## Calculus with Applications (10th Edition)

$$7$$
\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\left( {7 - x} \right)\ln \left( {1 - x} \right)}}{{{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} \left( {7 - x} \right)\ln \left( {1 - x} \right) = \left( {7 - 0} \right)\ln \left( {1 - 0} \right) = \left( 7 \right)\left( 0 \right) = 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 }}\left( {7 - x} \right)\ln \left( {1 - x} \right) \to {D_x}\left( {\left( {7 - x} \right)\ln \left( {1 - x} \right)} \right) \cr & {\text{By the product rule}} \cr & = \left( {7 - x} \right){D_x}\left( {\ln \left( {1 - x} \right)} \right) + \ln \left( {1 - x} \right){D_x}\left( {\left( {7 - x} \right)} \right) \cr & = \left( {7 - x} \right)\left( {\frac{{ - 1}}{{1 - x}}} \right) + \ln \left( {1 - x} \right)\left( { - 1} \right) \cr & = \frac{{7 - x}}{{x - 1}} - \ln \left( {1 - x} \right) \cr & \cr & {\text{for }}{e^{ - x}} - 1 \to {D_x}\left( {{e^{ - x}} - 1} \right) = - {e^{ - x}} \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\left( {7 - x} \right)\ln \left( {1 - x} \right)}}{{{e^{ - x}} - 1}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{7 - x}}{{x - 1}} - \ln \left( {1 - x} \right)}}{{ - {e^{ - x}}}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{{\frac{{7 - 0}}{{0 - 1}} - \ln \left( {1 - 0} \right)}}{{ - {e^{ - 0}}}} \cr & = \frac{{ - 7 - 0}}{{ - 1}} \cr & = 7 \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\left( {7 - x} \right)\ln \left( {1 - x} \right)}}{{{e^{ - x}} - 1}} = 7 \cr}