Answer
$$\lim _{x \rightarrow 0} \frac{x+x \cos x}{\sin x \cos x}=2$$
Work Step by Step
Given $$\lim _{x \rightarrow 0} \frac{x+x \cos x}{\sin x \cos x}$$
So, we get
\begin{aligned}L&=\lim _{x \rightarrow 0} \frac{x+x \cos x}{\sin x \cos x}\\
&=\lim _{x \rightarrow 0} \frac{x(1+ \cos x)}{\sin x \cos x}\\
&=\lim _{x \rightarrow 0}\left(\frac{x}{\sin x} \cdot \frac{1+\cos x}{\cos x}\right)\\
&=\lim _{x \rightarrow 0}\left(\frac{1}{\frac{\sin x}{x}} \cdot \frac{1+\cos x}{\cos x}\right)\\
&=\lim _{x \rightarrow 0} \frac{1}{\frac{\sin x}{x}}.\lim _{x \rightarrow 0} \frac{1+\cos x}{\cos x}\\
&=\frac{\lim _{x \rightarrow 0} 1}{\lim _{x \rightarrow 0} \frac{\sin x}{x}} \cdot \frac{\lim _{x \rightarrow 0}(1+\cos x)}{\lim _{x \rightarrow 0} \cos x}\\
&=\frac{1}{1} \cdot \frac{1+\cos 0}{1}\\
&=1 \cdot 2\\
&=2
\end{aligned}
