Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises: 19

Answer

$$\lim\limits_{x\to0^+}\tan^{-1}(1/x)=\frac{\pi}{2}$$

Work Step by Step

$$A=\lim\limits_{x\to0^+}\tan^{-1}(1/x)$$ Let $u=1/x$ As $x\to0^+$, $1/x$ approaches $\infty$. That means $u$ approaches $\infty$, too. Therefore, $$A=\lim\limits_{u\to\infty}\tan^{-1}u=\frac{\pi}{2}$$ *NOTES TO REMEMBER: $$\lim\limits_{x\to\infty}\tan^{-1}x=\frac{\pi}{2}$$ and $$\lim\limits_{x\to-\infty}\tan^{-1}x=-\frac{\pi}{2}$$
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