If $$f(x) = {{\sin ({e^{x - 2}} - 1)} \over {\ln (x - 1)}}$$, then $$\mathop {\lim }\limits_{x \to 2} $$ f(x) is equal to

Consider the function

$$f(x) = \left\{ {\matrix{ { - 2\sin x,} & {if} & {x \le - {\pi \over 2}} \cr {A\sin x + B,} & {if} & { - {\pi \over 2} < x < {\pi \over 2}} \cr {\cos x,} & {if} & {x \ge {\pi \over 2}} \cr } } \right.$$

which is continuous everywhere.

The value of A is

Consider the function

$$f(x) = \left\{ {\matrix{ { - 2\sin x,} & {if} & {x \le - {\pi \over 2}} \cr {A\sin x + B,} & {if} & { - {\pi \over 2} < x < {\pi \over 2}} \cr {\cos x,} & {if} & {x \ge {\pi \over 2}} \cr } } \right.$$

which is continuous everywhere.

The value of B is

Consider the following in respect of the function

$$f(x) = \left\{ \matrix{ 2 + x,x \ge 0 \hfill \cr 2 - x,x < 0 \hfill \cr} \right.$$.

I. $$\mathop {\lim }\limits_{x \to 1} f(x)$$ does not exist.

II. f(x) is differentiable at x = 0.

III. f(x) is continuous at x = 0.

Which of the above statements is/are correct?