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 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 25



Work Step by Step

Let $\lim\limits_{x \to (\dfrac{\pi}{2})^{-}}f(x)=\lim\limits_{x \to (\dfrac{\pi}{2})^{-}} (x-\dfrac{\pi}{2}) (\sec x)$ But $f((\dfrac{\pi}{2})^{-})=\dfrac{0}{0}$ This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$ This implies that $\lim\limits_{x \to (\dfrac{\pi}{2})^{-}}\dfrac{1}{(-\sin x)}=-\dfrac{1}{\sin (\dfrac{\pi}{2})}=-1$
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