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: 30

Answer

$${\text{ The limit does not exist}}$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {{x^2} + 5x + 9} }}{{x - 1}} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 1} \left( {\sqrt {{x^2} + 5x + 9} } \right) = \sqrt {{{\left( 1 \right)}^2} + 5\left( 1 \right) + 9} = \sqrt {15} \cr & \mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right) = 1 - 1 = 0 \cr & \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {{x^2} + 5x + 9} }}{{x - 1}} = \frac{{\sqrt {15} }}{0} \cr & {\text{Since the limits do not leadsto the indeterminate form }}\frac{0}{0}{\text{ or }}\frac{{ \pm \infty }}{{ \pm \infty }} \cr & {\text{we cannot apply the l'Hospital's rule}}. \cr & {\text{By l'Hospital' rule}}{\text{, the limit does not exist}} \cr} $$
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