Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 421: 97

Answer

$1$

Work Step by Step

$x$ is approaching 0 from the right, so it is positive $(x=\sqrt{x}\cdot\sqrt{x}$) Rewrite the limit $L=\displaystyle \lim_{x\rightarrow 0^{+}}\frac{(\tan^{-1}\sqrt{x})^{2}}{(\sqrt{x})^{2}}\cdot\lim_{x\rightarrow 0^{+}}\frac{1}{\sqrt{x+1}}=L_{1}\cdot L_{2}$ $L_{2}=1$ by direct substitution. $ L_{1}=[\displaystyle \lim_{x\rightarrow 0^{+}}\frac{\tan^{-1}\sqrt{x}}{\sqrt{x}}]^{2}\qquad$ The form of the limit is $\displaystyle \frac{0}{0}$. Apply L'Hospital's rule. From table 7-4, $\displaystyle \quad \frac{d(\tan^{-1}u)}{dx}=\frac{1}{1+u^{2}}\frac{du}{dx}$ $L_{1}=[\displaystyle \lim_{x\rightarrow 0^{+}}\frac{\frac{1}{1+(\sqrt{x})^{2}}\cdot\frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}]^{2}=[\lim_{x\rightarrow 0^{+}}\frac{1}{1+x}]^{2}=1$ $L=L_{1}\cdot L_{2}=1\cdot 1=1$
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