Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises - Page 138: 30



Work Step by Step

$$\lim\limits_{x\to\infty}\sqrt{x^2+1}$$$$=\sqrt{\lim\limits_{x\to\infty}(x^2)+1}$$ As $x\to\infty$, $x^2\to\infty$. Therefore, $\lim\limits_{x\to\infty}(x^2)=\infty$ Which means, $\sqrt{\lim\limits_{x\to\infty}(x^2)+1}=\sqrt{\infty+1}=\infty$ In conclusion, $\lim\limits_{x\to\infty}\sqrt{x^2+1}=\infty$
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