Chapter 12 - Sequences and Series - Chapter Review - Review Exercises - Page 662: 64

$$ - \frac{1}{3}$$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \left( {\frac{2}{{{x^3}}} + \frac{2}{{{x^2}}} + \frac{1}{x} - \frac{{2{e^x}}}{{{x^3}}}} \right) \cr & {\text{The common denominator is }}{x^3},{\text{ then}} \cr & \frac{2}{{{x^3}}} + \frac{2}{{{x^2}}} + \frac{1}{x} - \frac{{2{e^x}}}{{{x^3}}} = \frac{{2 + 2x + {x^2} - 2{e^x}}}{{{x^3}}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{2}{{{x^3}}} + \frac{2}{{{x^2}}} + \frac{1}{x} - \frac{{2{e^x}}}{{{x^3}}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{2 + 2x + {x^2} - 2{e^x}}}{{{x^3}}}} \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{{2 + 2x + {x^2} - 2{e^x}}}{{{x^3}}}} \right) = \frac{{2 + 2\left( 0 \right) + {{\left( 0 \right)}^2} - 2{e^0}}}{{{{\left( 0 \right)}^3}}} = \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 }}{e^x} - 1 - x \to {D_x}\left( {2 + 2x + {x^2} - 2{e^x}} \right) = 2 + 2x - 2{e^x} \cr & {\text{for }}{x^2} \to {D_x}\left( {{x^3}} \right) = 3{x^2} \cr & {\text{then}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2 + 2x - 2{e^x}}}{{3{x^2}}} = \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 }}{e^x} - 1 \to {D_x}\left( {2 + 2x - 2{e^x}} \right) = 2 - 2{e^x} \cr & {\text{for }}3{x^2} \to {D_x}\left( {3{x^2}} \right) = 6x \cr & {\text{then}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2 - 2{e^x}}}{{6x}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{2 - 2{e^x}}}{{6x}} = \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 }}{e^x} - 1 \to {D_x}\left( {2 - 2{e^x}} \right) = 2 - 2{e^x} \cr & {\text{for }}6x \to {D_x}\left( {6x} \right) = 6 \cr & {\text{then}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2{e^x}}}{6} \cr & = - \frac{{2{e^0}}}{6} = - \frac{1}{3} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{2}{{{x^3}}} + \frac{2}{{{x^2}}} + \frac{1}{x} - \frac{{2{e^x}}}{{{x^3}}}} \right) = - \frac{1}{3} \cr} $$