## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{\theta\to0}\frac{\cos\theta-1}{2\theta^2}=-\frac{1}{4}$$
$$A=\lim\limits_{\theta\to0}\frac{\cos\theta-1}{2\theta^2}$$ Try to create $\frac{\sin\theta}{\theta}$. In detail: $$A=\lim\limits_{\theta\to0}\frac{\cos\theta-1}{2\theta^2}\times\frac{\cos\theta+1}{\cos\theta+1}$$ $$A=\lim\limits_{\theta\to0}\frac{\cos^2\theta-1}{2\theta^2(\cos\theta+1)}$$ $$A=\lim\limits_{\theta\to0}\frac{-\sin^2\theta}{2\theta^2(\cos\theta+1)}$$ (since $\sin^2\theta+\cos^2\theta=1$ so $\cos^2\theta-1=-\sin^2\theta$) $$A=\lim\limits_{\theta\to0}\Bigg(-\frac{1}{2}\times\frac{\sin^2\theta}{\theta^2}\times\frac{1}{\cos\theta+1}\Bigg)$$ $$A=-\frac{1}{2}\Bigg(\lim\limits_{\theta\to0}(\frac{\sin\theta}{\theta})^2\times\lim\limits_{\theta\to0}\frac{1}{\cos\theta+1}\Bigg)$$ $$A=-\frac{1}{2}\Bigg(1^2\times\frac{1}{\cos0+1}\Bigg)$$ $$A=-\frac{1}{2}\times\frac{1}{2}=-\frac{1}{4}$$