$$\mathop {\lim }\limits_{x \to 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x}= $$

At $x=0, f(x)=\left\{\begin{array}{l}\frac{x}{|x|+2 x^2}, x \neq 0 \\ k, \quad x=0\end{array}\right.$ is

$$\mathop {\lim }\limits_{x \to 0} \frac{x \tan 4 x-2 x \tan 2 x}{(1-\cos 4 x)^2}= $$

Assertion (A) $f(x)=|x-a|+|x-b|$, is continuous on $\mathbf{R}$

Reason (R) $\frac{|x-\alpha|}{x-\alpha}$ is continuous at $x \in \mathbf{R}-\{\alpha\}$.

The correct option among the following is