## Calculus with Applications (10th Edition)

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