Since $\lim_{x\to 0} \frac{\sin(x)}{x}=1,\lim_{x\to 0} \frac{x}{\sin(x)}=1 $
$\lim_{x\to 0} \frac{\cos(x)}{\sin(x)}\cdot\frac{x}{x}=\lim_{x\to 0} \frac{x}{\sin(x)}\cdot\frac{\cos(x)}{x}=\lim_{x\to 0} \frac{x}{\sin(x)}\cdot\lim_{x\to 0}\frac{\cos(x)}{x}=1\cdot\lim_{x\to 0}\frac{\cos(x)}{x}$
1) I do not understand the transition from this step to this. $\lim_{x\to 0} \frac{x}{\sin(x)}\cdot\frac{\cos(x)}{x}=\lim_{x\to 0} \frac{x}{\sin(x)}\cdot\lim_{x\to 0}\frac{\cos(x)}{x}$
I thought that $\lim_{x\to a}A\cdot B= \lim_{x\to a}A\cdot\lim_{x\to a}B$ only if $\lim_{x\to a}A$ and $\lim_{x\to a}B$ exists? In this case, $\lim_{x\to 0}\frac{\cos(x)}{x}$ does not exist. How can we make this transition then?
I use this technique with trigonometric limits very frequently, but I do not understand the basis for this step.