## Calculus with Applications (10th Edition)

$$\frac{7}{4}$$
\eqalign{ & \mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - {x^2} - x - 2}}{{{x^2} - 4}} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}2{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 2} \left( {{x^3} - {x^2} - x - 2} \right) = {\left( 2 \right)^3} - {\left( 2 \right)^2} - \left( 2 \right) - 2 = 0 \cr & \mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4} \right) = {\left( 2 \right)^2} - 4 = 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}} \cr & = \mathop {\lim }\limits_{x \to 2} \frac{{{D_x}\left( {{x^3} - {x^2} - x - 2} \right)}}{{{D_x}\left( {{x^2} - 4} \right)}} \cr & = \mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - 2x - 1}}{{2x}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{{3{{\left( 2 \right)}^2} - 2\left( 2 \right) - 1}}{{2\left( 2 \right)}} \cr & = \frac{7}{4} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - {x^2} - x - 2}}{{{x^2} - 4}} = \frac{7}{4} \cr}