## Calculus with Applications (10th Edition)

$$2$$
\eqalign{ & \mathop {\lim }\limits_{x \to 0} \left( {\frac{{12{e^x}}}{{{x^3}}} - \frac{{12}}{{{x^3}}} - \frac{{12}}{{{x^2}}} - \frac{6}{x}} \right) \cr & {\text{The common denominator is }}{x^3},{\text{ then}} \cr & \frac{{12{e^x}}}{{{x^3}}} - \frac{{12}}{{{x^3}}} - \frac{{12}}{{{x^2}}} - \frac{6}{x} = \frac{{12{e^x} - 12 - 12x - 6{x^2}}}{{{x^3}}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{{12{e^x}}}{{{x^3}}} - \frac{{12}}{{{x^3}}} - \frac{{12}}{{{x^2}}} - \frac{6}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{12{e^x} - 12 - 12x - 6{x^2}}}{{{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{{12{e^x} - 12 - 12x - 6{x^2}}}{{{x^3}}}} \right) = \frac{{12{e^0} - 12 - 12\left( 0 \right) - 6{{\left( 0 \right)}^2}}}{{{0^3}}} = \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 }}12{e^x} - 12 - 12x - 6{x^2} \to {D_x}\left( {12{e^x} - 12 - 12x - 6{x^2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 12{e^x} - 12 - 12x \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} \left( {\frac{{12{e^x} - 12 - 12x}}{{3{x^2}}}} \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 }}12{e^x} - 12 - 12x \to {D_x}\left( {12{e^x} - 12 - 12x} \right) = 12{e^x} - 12 \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{{12{e^x} - 12}}{{6x}} = \frac{0}{0} \cr & {\text{l'Hospital's rule applies}}{\text{.}} \cr & {\text{for }}12{e^x} - 12 \to {D_x}\left( {12{e^x} - 12} \right) = 12{e^x} \cr & {\text{for }}6x \to {D_x}\left( {6x} \right) = 6 \cr & \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{12{e^x}}}{6} = \frac{{12{e^0}}}{6} = 2 \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{{12{e^x}}}{{{x^3}}} - \frac{{12}}{{{x^3}}} - \frac{{12}}{{{x^2}}} - \frac{6}{x}} \right) = 2 \cr}