Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 500: 37

Answer

\[\lim_{x\rightarrow 0^+}\frac{\tan^{-1} 2x}{\ln x}=0\]

Work Step by Step

Let \[l=\lim_{x\rightarrow 0^+}\frac{\tan^{-1} 2x}{\ln x}\] Which is $\frac{0}{0}$ form Using L'Hopital's rule \[\Rightarrow l=\lim_{x\rightarrow 0^+}\frac{(\tan^{-1} 2x)'}{(\ln x)'}\] \[\Rightarrow l=\lim_{x\rightarrow 0^+}\frac{\frac{2}{1+4x^2}}{\frac{1}{x}}\] \[\Rightarrow l=\lim_{x\rightarrow 0^+}\frac{2x}{1+4x^2}\] \[\Rightarrow l=\lim_{x\rightarrow 0^+}\frac{2x}{1+4x^2}=0\] Hence, \[\lim_{x\rightarrow 0^+}\frac{\tan^{-1} 2x}{\ln x}=0\]
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