Answer
$$\lim_{x\to0}\frac{x-x\cos x}{\sin^23x}=0$$
Work Step by Step
$$A=\lim_{x\to0}\frac{x-x\cos x}{\sin^23x}=\lim_{x\to0}\frac{x(1-\cos x)}{\sin^23x}$$
$$A=\lim_{x\to0}\frac{x}{\sin3x}\times\lim_{x\to0}\frac{1-\cos x}{\sin3x}=X\times Y$$
1) Consider $X$: $$X=\lim_{x\to0}\frac{x}{\sin 3x}=\Big(\lim_{x\to0}\frac{\sin3x}{x}\Big)^{-1}$$
Multiply both numerator and denominator by $3$:
$$X=\Big(\lim_{x\to0}\frac{3\sin3x}{3x}\Big)^{-1}=\Big(3\lim_{x\to0}\frac{\sin3x}{3x}\Big)^{-1}$$
Apply Theorem 7 here with $\theta=3x$: $$X=(3\times1)^{-1}=3^{-1}=\frac{1}{3}$$
2) Consider $Y$: $$Y=\lim_{x\to0}\frac{1-\cos x}{\sin3x}$$
$\sin3x$ can be written into $\sin(2x+x)$
That means $$\sin3x=\sin2x\cos x+\cos2x\sin x$$ $$\sin3x=(2\sin x\cos x)\cos x+(1-2\sin^2x)\sin x$$ $$\sin3x=2\sin x\cos^2x+\sin x-2\sin^3x$$ $$\sin3x=2\sin x(1-\sin^2x)+\sin x-2\sin^3x$$ $$\sin3x=2\sin x-2\sin^3x+\sin x-2\sin^3x$$ $$\sin3x=3\sin x-4\sin^3x$$
Therefore, $$Y=\lim_{x\to0}\frac{1-\cos x}{3\sin x-4\sin^3x}$$
Multiply both numerator and denominatory by $1+\cos x$:
$$Y=\lim_{x\to0}\frac{(1-\cos x)(1+\cos x)}{(3\sin x-4\sin^3x)(1+\cos x)}$$
$$Y=\lim_{x\to0}\frac{1-\cos^2x}{\sin x(3-4\sin^2x)(1+\cos x)}$$
$$Y=\lim_{x\to0}\frac{\sin^2x}{\sin x(3-4\sin^2x)(1+\cos x)}$$
$$Y=\lim_{x\to0}\frac{\sin x}{(3-4\sin^2x)(1+\cos x)}$$
$$Y=\frac{\sin0}{(3-4\sin^20)(1+\cos0)}$$
$$Y=\frac{0}{(3-4\times0)(1+1)}=0$$
In conclusion, $$A=X\times Y=\frac{1}{3}\times0=0$$
