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

$$\lim_{\theta\to0}\frac{1-\cos\theta}{\theta^2}=\frac{1}{2}$$
$$A=\lim_{\theta\to0}\frac{1-\cos\theta}{\theta^2}$$ Multiply both numerator and denominator with $1+\cos\theta$: $$A=\lim_{\theta\to0}\frac{(1-\cos\theta)(1+\cos\theta)}{\theta^2(1+\cos\theta)}$$ $$A=\lim_{\theta\to0}\frac{1-\cos^2\theta}{\theta^2(1+\cos\theta)}$$ Recall the identity: $1-\cos^2\theta=\sin^2\theta$: $$A=\lim_{\theta\to0}\frac{\sin^2\theta}{\theta^2(1+\cos\theta)}$$ $$A=\lim_{\theta\to0}\frac{\sin^2\theta}{\theta^2}\times\lim_{\theta\to0}\frac{1}{1+\cos\theta}$$ We have $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$ $$A=1^2\times\frac{1}{1+\cos0}$$ $$A=\frac{1}{1+1}=\frac{1}{2}$$