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: 38

Answer

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

Work Step by Step

The solve this limit follow the steps below $$\lim_{x\to\infty}\frac{4x^3-2x^2+6}{\pi x^3+4}=\lim_{x\to\infty}\frac{x^3(4-\frac{2x^2}{x^3}+\frac{6}{x^3})}{x^3(\pi+\frac{4}{x^3})}=\lim_{x\to\infty}\frac{4-\frac{2}{x}+\frac{6}{x^3}}{\pi+\frac{4}{x^3}}=\left[\frac{4-\frac{2}{\infty}+\frac{6}{\infty^3}}{\pi+\frac{4}{\infty^3}}\right]=\left[\frac{4-0+0}{\pi+0}\right]=\frac{4}{\pi}.$$
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