Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 12 - Sequences and Series - 12.7 L'Hospital's Rule - 12.7 Exercises - Page 659: 2

Answer

$$\frac{{22}}{3}$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to 3} \frac{{{x^3} + {x^2} - 11x - 3}}{{{x^2} - 3x}} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}3{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 3} \left( {{x^3} + {x^2} - 11x - 3} \right) = {\left( 3 \right)^3} + {\left( 3 \right)^2} - 11\left( 1 \right) - 3 = 0 \cr & \mathop {\lim }\limits_{x \to 3} \left( {{x^2} - 3x} \right) = {\left( 3 \right)^2} - \left( 3 \right)\left( 3 \right) = 0 \cr & \cr & {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr & {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr & {\text{denominator}} \cr & = \mathop {\lim }\limits_{x \to 3} \frac{{{D_x}\left( {{x^3} + {x^2} - 11x - 3} \right)}}{{{D_x}\left( {{x^2} - 3x} \right)}} \cr & = \mathop {\lim }\limits_{x \to 3} \frac{{3{x^2} + 2x - 11}}{{2x - 3}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{{3{{\left( 3 \right)}^2} + 2\left( 3 \right) - 11}}{{2\left( 3 \right) - 1}} \cr & = \frac{{22}}{3} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 3} \frac{{{x^3} + {x^2} - 11x - 3}}{{{x^2} - 3x}} = \frac{{22}}{3} \cr} $$
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