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: 28



Work Step by Step

We have $$ \lim _{x \rightarrow 0} \left(\cot x -\frac{1}{x}\right) =\infty-\infty $$ is an intermediate form, then we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow 0} \frac{x\cot x -1}{x}=\lim _{x \rightarrow 0} \frac{\cot x-x\csc^2 x }{1}=-0\times\infty+\infty .$$ Again we can apply L’Hôpital’s Rule $$ \lim _{x \rightarrow 0} ( \cot x-x\csc^2 x) = \lim _{x \rightarrow 0} \frac{\cos x \sin x-x}{\sin^2 x} = \lim _{x \rightarrow 0} \frac{- \sin^2 x+\cos^2 x-1}{2\sin x\cos x} =0 .$$
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