University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 248: 26



Work Step by Step

Consider: $\lim\limits_{x \to (\dfrac{\pi}{2})^{-}}f(x)=\lim\limits_{x \to (\dfrac{\pi}{2})^{-}} (x-\dfrac{\pi}{2}) \tan x$ or, $\lim\limits_{x \to (\dfrac{\pi}{2})^{-}} (x-\dfrac{\pi}{2}) \tan x=\lim\limits_{x \to (\dfrac{\pi}{2})^{-}} (x-\dfrac{\pi}{2}) \dfrac{\sin x}{\cos x}$ We need to check that the limit has an indeterminate form. Thus, $f((\dfrac{\pi}{2})^{-})=\dfrac{0}{0}$ The limit shows an indeterminate form. Thus, apply L-Hospital's rule: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ Then $\lim\limits_{x \to (\dfrac{\pi}{2})^{-}}\dfrac{-\sin x+(\pi/2 -x) \cos x}{-\sin x}=\dfrac{-\sin (\pi/2)+(\pi/2 -\pi/2) \cos (\pi/2)}{-\sin (\pi/2)}=\dfrac{-1}{-1}=1$
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