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: 40

Answer

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

Work Step by Step

$$A=\lim\limits_{x\to0^+}\tan^{-1}(\ln x)$$$$A=\lim\limits_{x\to0^+}\arctan(\ln x)$$ Let $t=\ln x$ As $x\to{0^+}$, $\ln x$ approaches $-\infty$. Therefore $t\to{-\infty}$ $$A=\lim\limits_{t\to-\infty}\arctan t$$$$A=\frac{-\pi}{2}$$ *NOTES TO REMEMBER: $\lim\limits_{x\to-\infty}\arctan x=\frac{-\pi}{2}$
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