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: 54

Answer

The solution is $$\lim_{x\to\infty}(x-\sqrt{x^2+4x})=-2.$$

Work Step by Step

To solve this limit follow the steps below $$\lim_{x\to\infty}(x-\sqrt{x^2+4x})=\lim_{x\to\infty}(x-\sqrt{x^2+4x})\cdot\frac{x+\sqrt{x^2+4x}}{x+\sqrt{x^2+4x}}=\lim_{x\to\infty}\frac{x^2-\sqrt{x^2+4x}^2}{x+\sqrt{x^2+4x}}=\lim_{x\to\infty}\frac{x^2-x^2-4x}{x+\sqrt{x^2+4x}}=\lim_{x\to\infty}\frac{-4x}{x+\sqrt{x^2+4x}}=\lim_{x\to\infty}\frac{-4x}{x\left(1+\sqrt{1+\frac{4}{x}}\right)}=\lim_{x\to\infty}\frac{-4}{1+\sqrt{1+\frac{4}{x}}}=\left[\frac{-4}{1+\sqrt{1+\frac{4}{\infty}}}\right]=\left[\frac{-4}{1+\sqrt{1+0}}\right]=-2.$$
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