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: 52

Answer

The solution is $$\lim_{x\to\infty}(x-\sqrt{x^2+1})=0.$$

Work Step by Step

To calculate this limit follow the steps below $$\lim_{x\to\infty}(x-\sqrt{x^2+1})=\lim_{x\to\infty}(x-\sqrt{x^2+1})\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}}=\lim_{x\to\infty}\frac{x^2-\sqrt{x^2+1}^2}{x+\sqrt{x^2+1}}=\lim_{x\to\infty}\frac{x^2-x^2-1}{x+\sqrt{x^2+1}}=\lim_{x\to\infty}\frac{-1}{x+\sqrt{x^2+1}}=\left[\frac{-1}{\infty}\right]=0.$$
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