Answer
$$\lim_{x\to0}\frac{x+x\cos x}{\sin x\cos x}=2$$
Work Step by Step
$$A=\lim_{x\to0}\frac{x+x\cos x}{\sin x\cos x}$$
Multiply both numerator and denominator by $2$, we have
$$A=\lim_{x\to0}\frac{2(x+x\cos x)}{2\sin x\cos x}$$
$$A=2\lim_{x\to0}\frac{x(1+\cos x)}{\sin2x}$$
$$A=2\lim_{x\to0}\frac{x}{\sin2x}\lim_{x\to0}(1+\cos x)$$
$$A=2\lim_{x\to0}\Big(\frac{\sin2x}{x}\Big)^{-1}(1+\cos0)$$
$$A=2\Big(\lim_{x\to0}\frac{\sin2x}{x}\Big)^{-1}(1+1)=2\Big(\lim_{x\to0}\frac{\sin2x}{x}\Big)^{-1}\times2$$
$$A=4\Big(\lim_{x\to0}\frac{\sin2x}{x}\Big)^{-1}=4X^{-1}$$
Considering $X$: $$X=\lim_{x\to0}\frac{\sin2x}{x}$$
Multiply both numerator and denominator by $2$: $$X=\lim_{x\to0}\frac{2\sin2x}{2x}=2\lim_{x\to0}\frac{\sin2x}{2x}$$
Apply Theorem 7 with $\theta=2x$: $$X=2\times1=2$$
Therefore, $$A=4X^{-1}=4\times(2)^{-1}=4\times\frac{1}{2}=2$$
