Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 41

Answer

The solution is $$\lim_{x\to\infty}\frac{\ln(3x+5)}{\ln(7x+3)+1}=1.$$

Work Step by Step

To solve this limit follow the steps below. "LR" will stand for "Apply L'Hopital's rule." $$\lim_{x\to\infty}\frac{\ln(3x+5)}{\ln(7x+3)+1}=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(\ln(3x+5))'}{(\ln(7x+3)+1)'}=\lim_{x\to\infty}\frac{\frac{1}{3x+5}(3x+5)'}{\frac{1}{7x+3}(7x+3)'}=\lim_{x\to\infty}\frac{\frac{3}{3x+5}}{\frac{7}{7x+3}}=\lim_{x\to\infty}\frac{3(7x+3)}{7(3x+5)}=\lim_{x\to\infty}\frac{3x\left(7+\frac{3}{x}\right)}{7x\left(3+\frac{5}{x}\right)}=\lim_{x\to\infty}\frac{3\left(7+\frac{3}{x}\right)}{7\left(3+\frac{5}{x}\right)}=\left[\frac{3\left(7+\frac{3}{\infty}\right)}{7\left(3+\frac{5}{\infty}\right)}\right]=\left[\frac{3(7+0)}{7(3+0)}\right]=1.$$
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