## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 17

#### Answer

$-2$

#### Work Step by Step

Consider: $\lim\limits_{\theta \to \dfrac{\pi}{2}}f(x)=\lim\limits_{ \theta \to \dfrac{\pi}{2}}\dfrac{2 \theta -\pi}{\cos (2\pi-\theta)}$ Need to check that the limit has an indeterminate form. Thus, $f(\dfrac{\pi}{2})=\dfrac{2 (\dfrac{\pi}{2}) -\pi}{\cos (2\pi-(\dfrac{\pi}{2}))}=\dfrac{0}{0}$ Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ This implies: $\lim\limits_{\theta \to \dfrac{\pi}{2}}\dfrac{\cos x -1}{3x^2}=\dfrac{0}{0}$ Again apply L-Hospital's rule: we get $\lim\limits_{\theta \to \dfrac{\pi}{2}}\dfrac{2}{\sin (2\pi -\theta)}=\dfrac{2}{\sin (2\pi -\dfrac{\pi}{2})}=-2$

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