## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 12 - Sequences and Series - 12.7 L'Hospital's Rule - 12.7 Exercises - Page 659: 24

#### Answer

$$\frac{1}{{12}}$$

#### Work Step by Step

\eqalign{ & \mathop {\lim }\limits_{x \to 5} \frac{{\sqrt {{x^2} + 11} - 6}}{{{x^2} - 25}} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}5{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 5} \left( {\sqrt {{x^2} + 11} - 6} \right) = \sqrt {{5^2} + 11} - 6 = 6 - 6 = 0 \cr & \mathop {\lim }\limits_{x \to 5} \left( {{x^2} - 25} \right) = {5^2} - 25 = 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}}{\text{}} \cr & {\text{for }}\sqrt {{x^2} + 11} - 6 \to {D_x}\left( {\sqrt {{x^2} + 11} - 6} \right) = \frac{{2x}}{{2\sqrt {{x^2} + 11} }} - 0 = \frac{x}{{\sqrt {{x^2} + 11} }} \cr & {\text{for }}{x^2} - 25 \to {D_x}\left( {{x^2} - 25} \right) = 2x \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 5} \frac{{\sqrt {{x^2} + 11} - 6}}{{{x^2} - 25}} = \mathop {\lim }\limits_{x \to 5} \frac{{\frac{x}{{\sqrt {{x^2} + 11} }}}}{{2x}} = \mathop {\lim }\limits_{x \to 5} \frac{1}{{2\sqrt {{x^2} + 11} }} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{1}{{2\sqrt {{{\left( 5 \right)}^2} + 11} }} \cr & = \frac{1}{{2\sqrt {36} }} \cr & = \frac{1}{{12}} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 5} \frac{{\sqrt {{x^2} + 11} - 6}}{{{x^2} - 25}} = \frac{1}{{12}} \cr}

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