Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.7 L'Hôpital's Rule - Exercises - Page 366: 37



Work Step by Step

We have $$ \lim _{x \rightarrow 1}\tan (\frac{\pi x}{2})\ln x=\lim _{x \rightarrow 1}\frac{\sin(\frac{\pi x}{2})\ln x}{\cos\frac{\pi x}{2} }=\frac{0}{0} $$ which is an intermediate form, so we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow 1}\frac{\sin(\frac{\pi x}{2})\ln x}{\cos\frac{\pi x}{2} }= \lim _{x \rightarrow 1}\frac{\frac{1}{x}\sin(\frac{\pi x}{2})+\frac{\pi}{2}\cos(\frac{\pi x}{2}) \ln x}{-\frac{\pi}{2}\sin\frac{\pi x}{2} }\\ =-\frac{2}{\pi}. $$
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