## Calculus: Early Transcendentals 8th Edition

$$\lim_{\theta\to\pi/2}\frac{1-\sin\theta}{1+\cos2\theta}=\frac{1}{4}$$
$$A=\lim_{\theta\to\pi/2}\frac{1-\sin\theta}{1+\cos2\theta}$$ We can actually apply both methods here, as will be shown below. 1) Method 1: Elementary method Remember that in trigonometry, $1$ can also be written as $1=\sin^2\theta+\cos^2\theta$. Also, $\cos2\theta=\cos^2\theta-\sin^2\theta.$ $$A=\lim_{\theta\to\pi/2}\frac{1-\sin\theta}{(\sin^2\theta+\cos^2\theta)+(\cos^2\theta-\sin^2\theta)}$$ $$A=\lim_{\theta\to\pi/2}\frac{1-\sin\theta}{2\cos^2\theta}$$ Now we again apply $\sin^2\theta+\cos^2\theta=1$, with a little twist $\cos^2\theta=1-\sin^2\theta$. $$A=\lim_{\theta\to\pi/2}\frac{1-\sin\theta}{2(1-\sin^2\theta)}$$ $$A=\lim_{\theta\to\pi/2}\frac{1-\sin\theta}{2(1-\sin\theta)(1+\sin\theta)}$$ $$A=\lim_{\theta\to\pi/2}\frac{1}{2(1+\sin\theta)}$$ $$A=\frac{1}{2(1+\sin(\pi/2))}$$ $$A=\frac{1}{2(1+1)}$$ $$A=\frac{1}{4}$$ 2) Method 2: L'Hospital's Rule Because $\lim_{\theta\to\pi/2}(1-\sin\theta)=1-\sin(\frac{\pi}{2})=1-1=0$ and $\lim_{\theta\to\pi/2}(1+\cos2\theta)=1+\cos(2\times\frac{\pi}{2})=1+\cos\pi=1-1=0,$ this limit is an indeterminate form of $\frac{0}{0}$, so as L'Hospital's Rule can be applied, we have $$A=\lim_{\theta\to\pi/2}\frac{\frac{d}{d\theta}(1-\sin\theta)}{\frac{d}{d\theta}(1+\cos2\theta)}$$ $$A=\lim_{\theta\to\pi/2}\frac{-\cos\theta}{-2\sin2\theta}$$ $$A=\lim_{\theta\to\pi/2}\frac{\cos\theta}{2\sin2\theta}.$$ For $\lim_{\theta\to\pi/2}(\cos\theta)=\cos(\frac{\pi}{2})=0$ and $\lim_{\theta\to\pi/2}(2\sin2\theta)=2\sin(2\frac{\pi}{2})=2\sin\pi=0,$ L'Hospital's Rule can be applied here: $$A=\lim_{\theta\to\pi/2}\frac{\frac{d}{d\theta}(\cos\theta)}{\frac{d}{d\theta}(2\sin2\theta)}$$ $$A=\lim_{\theta\to\pi/2}\frac{-\sin\theta}{4\cos2\theta}$$ $$A=\frac{-\sin(\pi/2)}{4\cos(2\times\pi/2)}$$ $$A=\frac{-1}{4\cos\pi}$$ $$A=\frac{-1}{4(-1)}=\frac{1}{4}$$