## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{\theta\to0}\frac{1-\cos\theta}{\sin2\theta}=0$$
$$A=\lim_{\theta\to0}\frac{1-\cos\theta}{\sin2\theta}=\lim_{\theta\to0}\frac{1-\cos\theta}{2\sin\theta\cos\theta}$$ Multiply both numerator and denominator by $(1+\cos\theta)$: $$A=\lim_{\theta\to0}\frac{(1-\cos\theta)(1+\cos\theta)}{2\sin\theta\cos\theta(1+\cos\theta)}$$ $$A=\lim_{\theta\to0}\frac{1-\cos^2\theta}{2\sin\theta\cos\theta(1+\cos\theta)}$$ Recall that $\sin^2\theta=1-\cos^2\theta$ $$A=\lim_{\theta\to0}\frac{\sin^2\theta}{2\sin\theta\cos\theta(1+\cos\theta)}$$ $$A=\lim_{\theta\to0}\frac{\sin\theta}{2\cos\theta(1+\cos\theta)}$$ $$A=\frac{\sin0}{2\cos0(1+\cos0)}$$ $$A=\frac{0}{2\times1(1+1)}=0$$