Chapter 12 - Sequences and Series - Chapter Review - Review Exercises - Page 662: 65

$$8$$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\ln \left( {1 - 4x} \right)}}{{{x^2}}} + \frac{4}{x}} \right) \cr & {\text{The common denominator is }}{x^2},{\text{ then}} \cr & \frac{{\ln \left( {1 - 4x} \right)}}{{{x^2}}} + \frac{4}{x} = \frac{{\ln \left( {1 - 4x} \right) + 4x}}{{{x^2}}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\ln \left( {1 - 4x} \right)}}{{{x^2}}} + \frac{4}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\ln \left( {1 - 4x} \right) + 4x}}{{{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{{\ln \left( {1 - 4x} \right) + 4x}}{{{x^2}}}} \right) = \frac{{\ln \left( 1 \right) + 0}}{{{0^2}}} = \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 }}\ln \left( {1 - 4x} \right) + 4x \to {D_x}\left( {\ln \left( {1 - 4x} \right) + 4x} \right) = \frac{{ - 4}}{{1 - 4x}} + 4 \cr & {\text{for }}{x^2} \to {D_x}\left( {{x^2}} \right) = 2x \cr & {\text{then}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ - 4}}{{1 - 4x}} + 4}}{{2x}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{{\frac{{ - 4}}{{1 - 4\left( 0 \right)}} + 4}}{{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 }}\frac{{ - 4}}{{1 - 4x}} + 4 \to {D_x}\left( {\frac{{ - 4}}{{1 - 4x}} + 4} \right) = - 4\left( {\frac{{ - 4}}{{{{\left( {1 - 4x} \right)}^2}}}} \right) = \frac{{16}}{{{{\left( {1 - 4x} \right)}^2}}} \cr & {\text{for }}2x \to {D_x}\left( {2x} \right) = 2 \cr & {\text{then}} \cr & = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\frac{{16}}{{{{\left( {1 - 4x} \right)}^2}}}}}{2}} \right) \cr & = \mathop {\lim }\limits_{x \to 0} \frac{8}{{{{\left( {1 - 4x} \right)}^2}}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{8}{{{{\left( {1 - 4\left( 0 \right)} \right)}^2}}} = 8 \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\ln \left( {1 - 4x} \right)}}{{{x^2}}} + \frac{4}{x}} \right) = 8 \cr} $$