If $f: \mathbf{R} \rightarrow \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}$ be defined by $f(x)=\left\{\begin{array}{cc}x+2, & x>0 \\ 2-x, & x \leq 0\end{array}\right.$ and $g(x)=\left\{\begin{array}{cc}x^2-2 x-2, & 1 \leq x<2 \\ x-7 & x \geq 2 \\ x+5, & x<1\end{array}\right.$ then $\lim _{x \rightarrow 0} g \circ f(x)$

Define $f: R \rightarrow R$ by $f(x)= \begin{cases}(x-a) \frac{e^{\frac{1}{(x-a)}}-1}{\frac{1}{(x-a)}}+1 & \text { for } x \neq a \\ 0, \quad \text { at } x=a\end{cases}$

Then which one of the following is true?

Let $f, g: \mathbf{R} \rightarrow \mathbf{R}$ be functions defined by

$$ f(x)=\left\{\begin{array}{cc} x \sin \left(\frac{1}{x}\right), & \text { for } x \neq 0 \\ 0, & \text { for } x=0 \end{array}\right. $$

and $g(x)=x f(x)$

Consider the following statements

(i) $f(x)$ is continuous at $x=0$ but not differentiable at $x=0$

(ii) $g(x)$ is differentiable at $x=0$, but $g^1(x)$ is not continuous at $x=0$

Then, which one of the following is true?

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