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: 43

Answer

The solution is $$\lim_{x\to\infty}\frac{x^2-\ln\frac{2}{x}}{3x^2+2x}=\frac{1}{3}.$$

Work Step by Step

We will solve this limit using L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule." $$\lim_{x\to\infty}\frac{x^2-\ln\frac{2}{x}}{3x^2+2x}=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(x^2-\ln\frac{2}{x})'}{(3x^2+2x)'}=\lim_{x\to\infty}\frac{2x-\frac{1}{2/x}(2/x)'}{6x+2}=\lim_{x\to\infty}\frac{2x-\frac{x}{2}\left(-\frac{2}{x^2}\right)}{6x+2}=\lim_{x\to\infty}\frac{2x+\frac{1}{x}}{6x+2}=\lim_{x\to\infty}\frac{x\left(2+\frac{1}{x^2}\right)}{x\left(6+\frac{2}{x}\right)}=\lim_{x\to\infty}\frac{2+\frac{1}{x^2}}{6+\frac{2}{x}}=\left[\frac{2+\frac{1}{\infty^2}}{6+\frac{2}{\infty}}\right]=\left[\frac{2+0}{6+0}\right]=\frac{1}{3}.$$
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