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

$$\lim_{x\to0}\frac{\sin(1-\cos x)}{x}=0$$
$$A=\lim_{x\to0}\frac{\sin(1-\cos x)}{x}$$ $$A=\lim_{x\to0}\frac{\sin(1-\cos x)}{1-\cos x}\times\lim_{x\to0}\frac{1-\cos x}{x}$$ - For all $x\ne0$, $(1-\cos x)\ne0$. So $\lim_{x\to0}\frac{\sin(1-\cos x)}{1-\cos x}=1$ $$A=1\times\lim_{x\to0}\frac{1-\cos x}{x}$$ $$A=\lim_{x\to0}\frac{1-\cos x}{x}$$ Multiply both numerator and denominator by $1+\cos x$: $$A=\lim_{x\to0}\frac{(1-\cos x)(1+\cos x)}{x(1+\cos x)}$$ $$A=\lim_{x\to0}\frac{1-\cos^2 x}{x(1+\cos x)}$$ $$A=\lim_{x\to0}\frac{\sin^2x}{x(1+\cos x)}$$ $$A=\lim_{x\to0}\frac{\sin x}{x}\times\lim_{x\to0}\frac{\sin x}{1+\cos x}$$ $$A=1\times\Big(\frac{\sin0}{1+\cos0}\Big)$$ $$A=\frac{0}{1+1}=0$$