# Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 19

$\dfrac{1}{4}$

#### Work Step by Step

Consider: $\lim\limits_{\theta \to \dfrac{\pi}{2}}f(x)=\lim\limits_{ \theta \to \dfrac{\pi}{2}}\dfrac{1-\sin \theta }{1+\cos (2\theta)}$ Need to check that the limit has an indeterminate form. Thus, $f(\dfrac{\pi}{2})=\dfrac{1-\sin (\dfrac{\pi}{2}) }{1+\cos (2 (\dfrac{\pi}{2}))}=\dfrac{1-1}{1 +(-1)}=\dfrac{0}{0}$ 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: $\lim\limits_{\theta \to \dfrac{\pi}{2}}\dfrac{- \cos \theta}{-2 \sin 2\theta}=\dfrac{- \cos \dfrac{\pi}{2}}{-2 \sin (2 \dfrac{\pi}{2})}=\dfrac{0}{0}$ Now, again apply L-Hospital's rule . $\lim\limits_{\theta \to \dfrac{\pi}{2}}\dfrac{ \sin \theta}{-4 \cos 2\theta}=\dfrac{ \sin \dfrac{\pi}{2}}{-4 \cos 2 (\dfrac{\pi}{2})}=\dfrac{1}{4}$

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