Calculus with Applications (10th Edition)

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

Chapter 12 - Sequences and Series - Chapter Review - Review Exercises - Page 662: 54

Answer

$$3$$

Work Step by Step

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