Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.7 Exercises - Page 565: 83

Answer

$a)$ After taking the derivative multiple times, the function repeats itself. $b)$ $1$ $c)$ see picture

Work Step by Step

$a)$$\lim\limits_{x \to \infty}\frac{x}{\sqrt (x^{2}+1)}$=$\frac{\infty}{\infty}$ $\lim\limits_{x \to \infty}\frac{1}{\frac{1}{2}(x^{2}+1)^{-\frac{1}{2}}(2x)}$ $\lim\limits_{x \to \infty}\frac{\sqrt (x^{2}+1)}{x}$=$\frac{\infty}{\infty}$ $\lim\limits_{x \to \infty}\frac{\frac{1}{2}(x^{2}+1)^{-\frac{1}{2}}(2x)}{1}$ $\lim\limits_{x \to \infty}\frac{x}{\sqrt (x^{2}+1)}$ The limit repeats itself. $b)$ $\lim\limits_{x \to \infty}\frac{x}{\sqrt (x^{2}+1)}$ $\lim\limits_{x \to \infty}\frac{x(\frac{1}{x})}{\sqrt (x^{2}+1)(\frac{1}{\sqrt x^{2}})}$ $\lim\limits_{x \to \infty}\frac{1}{\sqrt (1+\frac{1}{x^{2}})}$=$\frac{1}{\sqrt 1}$=$1$ $c)$ see picture
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