## Calculus with Applications (10th Edition)

$$2$$
\eqalign{ & \mathop {\lim }\limits_{x \to 0} \left( {\frac{2}{x} - \frac{{\ln \left( {1 + 2x} \right)}}{{{x^2}}}} \right) \cr & {\text{The common denominator is }}{x^2},{\text{ then}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{{2x - \ln \left( {1 + 2x} \right)}}{{{x^2}}}} \right) \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( {\frac{{2x - \ln \left( {1 + 2x} \right)}}{{{x^2}}}} \right) = \frac{{2\left( 0 \right) - \ln \left( {1 + 2\left( 0 \right)} \right)}}{{{{\left( 0 \right)}^2}}} = \frac{0}{0} \cr & \cr & {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr & {\text{l'Hospital's rule applies}}{\text{. then differentiating the numerator and}} \cr & {\text{denominator}}{\text{.}} \cr & {\text{for }}2x - \ln \left( {1 + 2x} \right) \to {D_x}\left( {2x - \ln \left( {1 + 2x} \right)} \right) = 2 - \frac{2}{{1 + 2x}} \cr & {\text{for }}\left( {{x^2}} \right)\ln x \to {D_x}\left( {{x^2}} \right) = 2x \cr & {\text{then}} \cr & = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{2 - \frac{2}{{1 + 2x}}}}{{2x}}} \right) \cr & = \frac{{2 - \frac{2}{{1 + 2\left( 0 \right)}}}}{{2\left( 0 \right)}} = \frac{0}{0} \cr & {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr & {\text{l'Hospital's rule applies}}{\text{.}} \cr & {\text{for }}2 - \frac{2}{{1 + 2x}} \to {D_x}\left( {2 - \frac{2}{{1 + 2x}}} \right) = \frac{4}{{{{\left( {1 + 2x} \right)}^2}}} \cr & {\text{for }}\left( {2x} \right)\ln x \to {D_x}\left( {2x} \right) = 2 \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{{2 - \frac{2}{{1 + 2x}}}}{{2x}}} \right) = \mathop {\frac{1}{2}\lim }\limits_{x \to 0} \left( {\frac{4}{{{{\left( {1 + 2x} \right)}^2}}}} \right) \cr & {\text{Find the limit}} \cr & = \frac{1}{2}\left( {\frac{4}{{{{\left( {1 + 2\left( 0 \right)} \right)}^2}}}} \right) \cr & = 2 \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{2}{x} - \frac{{\ln \left( {1 + 2x} \right)}}{{{x^2}}}} \right) = 2 \cr}