University Calculus: Early Transcendentals (3rd Edition)

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