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 - Page 307: 37


The solution is $$\lim_{x\to\infty}\frac{3x^4-x^2}{6x^4+12}=\frac{1}{2}.$$

Work Step by Step

To solve this limit follow the steps below: $$\lim_{x\to\infty}\frac{3x^4-x^2}{6x^4+12}=\lim_{x\to\infty}\frac{x^4(3-\frac{x^2}{x^4})}{x^4(6+\frac{12}{x^4})}=\lim_{x\to\infty}\frac{3-\frac{1}{x^2}}{6+\frac{12}{x^4}}=\left[\frac{3-\frac{1}{\infty^2}}{6+\frac{12}{\infty^4}}\right]=\frac{3}{6}=\frac{1}{2}.$$
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